Analyse polynomial curves, stationary points and extrema

Which first step finds the stationary inputs of the displayed function?

Try all five inputs and compare function height with tangent gradient

Which quantity identifies the two stationary inputs?

Analysing a curve with its derivative

The derivative gives the stationary inputs and the direction in which the function moves. Evaluate the original function at the resulting inputs to obtain the points on the curve

1. 1Find stationary inputs A stationary input is found by solving the derivative equation equal to zero \(\displaystyle f'(x)=0\quad\Rightarrow\quad x=a\)
2. 2Read the sign change A negative-to-positive derivative change gives a local minimum, while a positive-to-negative change gives a local maximum \(\displaystyle -\;\to\;+\;\Rightarrow\;\text{minimum},\quad +\;\to\;-\;\Rightarrow\;\text{maximum}\)
3. 3Use derivative signs for intervals If the derivative is positive throughout an interval, the function is increasing there. If it is negative throughout an interval, the function is decreasing there \(\displaystyle f'(x)>0\text{ for all }x\in I\Rightarrow f\text{ increasing on }I,\quad f'(x)<0\text{ for all }x\in I\Rightarrow f\text{ decreasing on }I\)
4. 4Evaluate the original function Use the original function to find the ordinate of each stationary point and any endpoint \(\displaystyle f(-1)=(-1)^{3}-3(-1)=2\)

Given

\(\displaystyle f'(x)=3x^{2}-3=0\)

Solve the derivative equation and enter the smaller stationary input

Given

\(\displaystyle f(-1)=(-1)^{3}-3\times(-1)\)

Evaluate the original function at the smaller stationary input

Which equation finds stationary inputs rather than x-intercepts?

State the interval convention

These questions request open derivative-sign intervals separated at every derivative zero. For x³, the derivative is positive on both (−∞,0) and (0,∞); the zero itself is omitted from that report. The original function can still be strictly increasing across zero

1. 1Report the requested derivative-sign intervals Use open intervals split at every derivative zero, even without a sign change. This reporting convention does not claim the whole function cannot be strictly increasing across a stationary input

For f(x) = x³, which report follows the requested split-at-zeros convention?

Recover a quadratic from a stationary point

For f(x) = 2x² + bx + c with stationary point (3,5), use f′(3) = 0 and f(3) = 5. The first equation determines b; the second determines c.

1. 1A stationary point supplies two different conditions The point (h,k) gives f(h) = k. Its horizontal tangent gives f′(h) = 0. Use both to recover unknown coefficients

Given

\(\displaystyle 12+b=0\)

Find b from the stationary condition

Given

\(\displaystyle 18+3(-12)+c=5\)

Use the given point to find c after b = −12.

Two stationary inputs recover two derivative coefficients

For f(x) = x³ + bx² + cx + d, stationary inputs −1 and 2 give 3 − 2b + c = 0 and 12 + 4b + c = 0. Subtract to solve for b, substitute for c, then use a given function value for d.

1. 1A stationary point supplies two different conditions The point (h,k) gives f(h) = k. Its horizontal tangent gives f′(h) = 0. Use both to recover unknown coefficients

Given

\(\displaystyle 9+6b=0\)

Find b from the two stationary conditions

Given

\(\displaystyle 3-2(-3/2)+c=0\)

Find c using the recovered b.

Inflection need not mean stationary

For x³, the derivative 3x² turns at zero and has value zero there: the original curve has a stationary inflection. For x³ + x, the derivative 3x² + 1 still turns at zero but has value 1: the inflection is nonstationary

1. 1Inflection concerns changing concavity A turning point of the derivative graph marks a change in the original curve’s slope trend. That inflection can have zero or nonzero tangent gradient

For f(x) = x³ + x at zero, which classification is correct?

Given

\(\displaystyle f(x)=2(x-1)^3+4(x-1)-3\)

Find the tangent gradient at the inflection input x = 1.

An unbounded domain can have a global extremum

For x⁴ + 2, every value is at least 2 and equality occurs only at zero. There is a global minimum even though the domain is unbounded. The function grows without bound, so it has no global maximum

1. 1Global extrema use the whole domain For these polynomial models, compare all included stationary candidates and included boundary points. On an unbounded domain, also check end behaviour; a local extremum need not be global

On the whole real domain, what are the global extrema of x⁴ + 2?

Respect open and closed endpoints

For f(x) = x² + 1 on (0,2], the function increases throughout the domain. Its largest value occurs at the included endpoint 2. Values approach 1 near the excluded input zero, but never equal 1.

1. 1An approached bound need not be attained A missing endpoint may provide a limiting value but not a minimum or maximum. State an infimum or supremum separately when that value is never reached

Given

\(\displaystyle f(x)=x^2+1,\quad0<x\le2\)

Find the attained maximum value

Function and domain

\(\displaystyle f(x)=x^2+1,\quad 0<x\le2\)

What should be reported at the excluded left endpoint in this example?

From features to a complete hand sketch

Draw each complete labelled curve on paper or a blank drawing surface

1. 1Prepare the whole drawing State the domain and range. Label axes and scale; find intercepts, endpoints, asymptotes and exact key points before joining anything
2. 2Show the correct connections Draw every allowed branch with the correct direction and shape. Use open/closed endpoints, leave gaps at excluded inputs, and show end behaviour. A handful of correct dots is not a complete sketch
3. 3Check the mathematics Substitute labelled points, check signs between roots, and compare both ends of each branch. Use derivative/concavity information when available. A plotted model is feedback, not a drawing submitted by you
4. 4Written practice (ungraded) Compare your drawing with every listed feature and annotate corrections

Draw the curve from derivative evidence

Intercepts are x=−√3,0,√3. The derivative 3(x−1)(x+1) has signs +,−,+, giving a local maximum (−1,2) and local minimum (1,−2). The second derivative 6x changes sign at (0,0), a nonstationary inflection with slope −3. Domain/range R; the left end falls and the right end rises. Join one smooth curve consistent with all these facts

Rule

\(\displaystyle f(x)=x^3-3x\)

Find the tangent gradient at the inflection input x = 0.

Your complete hand-sketch task

Sketch g(x)=(x−1)³−(x−1), including all intercepts, stationary points, inflection, concavity and end behaviour. Then sketch h(x)=(x−2)³+1 separately and explain why its horizontal tangent at (2,1) is not a maximum or minimum

1. 1Prepare the whole drawing State the domain and range. Label axes and scale; find intercepts, endpoints, asymptotes and exact key points before joining anything
2. 2Show the correct connections Draw every allowed branch with the correct direction and shape. Use open/closed endpoints, leave gaps at excluded inputs, and show end behaviour. A handful of correct dots is not a complete sketch
3. 3Check the mathematics Substitute labelled points, check signs between roots, and compare both ends of each branch. Use derivative/concavity information when available. A plotted model is feedback, not a drawing submitted by you
4. 4Written practice (ungraded) Compare your drawing with every listed feature and annotate corrections

Compare the complete drawing

For g, zeros are 0,1,2; stationary inputs 1±1/√3 give ordinates ±2/(3√3), with the maximum at the smaller input. Inflection (1,0) has gradient −1. For h, the stationary inflection is (2,1), derivative 3(x−2)² is positive on both sides, x-intercept 1 and y-intercept −7. Both cubics have domain/range R and rising overall end behaviour

A quartic can touch both intercepts

This even function is symmetric about the y-axis. Zeros at ±1 are double roots, so the curve touches the axis at the minima (−1,0),(1,0). The local maximum is (0,1); both ends rise. The derivative 4x(x²−1) has signs −,+,−,+. Inflection inputs ±1/√3 have ordinate 4/9. Domain R; range [0,∞); no maximum on the whole domain

Rule

\(\displaystyle f(x)=(x^2-1)^2\)

Calculate the ordinate at an inflection input x = 1/√3.

Your complete hand-sketch task

Sketch the curve using derivative signs. Label roots with their crossing/touching behaviour, stationary points, inflection points, symmetry, range and both ends. Explain the difference between the local maximum and a global maximum

1. 1Prepare the whole drawing State the domain and range. Label axes and scale; find intercepts, endpoints, asymptotes and exact key points before joining anything
2. 2Show the correct connections Draw every allowed branch with the correct direction and shape. Use open/closed endpoints, leave gaps at excluded inputs, and show end behaviour. A handful of correct dots is not a complete sketch
3. 3Check the mathematics Substitute labelled points, check signs between roots, and compare both ends of each branch. Use derivative/concavity information when available. A plotted model is feedback, not a drawing submitted by you
4. 4Written practice (ungraded) Compare your drawing with every listed feature and annotate corrections

Compare the complete drawing

Double roots/minima at (−2,0),(2,0); local maximum (0,16). Derivative 4x(x²−4), signs −,+,−,+. Inflections at x=±2/√3, y=64/9. Domain R, range [0,∞), symmetry about y-axis. Both tails rise without bound, so the local maximum value 16 is not a global maximum

Draw both pieces with the correct join

Draw the parabola only on [−2,1), with a filled endpoint (−2,4) and an open point (1,1). Draw the line only on [1,3], with filled points (1,3),(3,7). The included value f(1)=3 is different from the left limit 1. The graph has a jump. Domain [−2,3], range [0,7]; the y-intercept and only root are (0,0)

Rule

\(\displaystyle f(x)=\begin{cases}x^2&-2\le x<1\\2x+1&1\le x\le3\end{cases}\)

Which ordinate is the actual function value f(1)?

Your complete hand-sketch task

Sketch g(x)=x² for −1≤x≤1 and g(x)=2x−1 for 1<x≤3. State the complete domain/range, label endpoints and the join, and decide whether the function and tangent are continuous at 1. Also sketch the composite q(x)=(2x−1)²−4 using its vertex and intercepts

1. 1Prepare the whole drawing State the domain and range. Label axes and scale; find intercepts, endpoints, asymptotes and exact key points before joining anything
2. 2Show the correct connections Draw every allowed branch with the correct direction and shape. Use open/closed endpoints, leave gaps at excluded inputs, and show end behaviour. A handful of correct dots is not a complete sketch
3. 3Check the mathematics Substitute labelled points, check signs between roots, and compare both ends of each branch. Use derivative/concavity information when available. A plotted model is feedback, not a drawing submitted by you
4. 4Written practice (ungraded) Compare your drawing with every listed feature and annotate corrections

Compare the complete drawing

For g, the join is the filled point (1,1); both one-sided slopes are 2 and both values approach 1, so it joins smoothly. Domain [−1,3], range [0,5], minimum (0,0), maximum (3,5). For q, vertex (1/2,−4), roots −1/2 and 3/2, y-intercept −3, domain R and range [−4,∞). Use a smooth upward parabola, not the individual graphs of the inner and outer functions

Turn examples into a parameter classification

An intersection count can change when two roots merge, a root reaches an excluded boundary, or the graph shape changes. Derive the possible transition values exactly; a slider or a finite collection of examples cannot prove the complete classification

1. 1Keep the parameter free A chosen parameter value gives one example, not the classification of a whole family. Derive the critical parameter values and test each interval and equality case
2. 2Count distinct allowed inputs After substituting u = x², keep only u ≥ 0. A positive u gives two real inputs, zero gives one, and a repeated u is not counted twice. Apply any original domain restriction afterwards \(\displaystyle u>0\Rightarrow x=\pm\sqrt u;\quad u=0\Rightarrow x=0\)
3. 3Compare every relevant height A minimum says whether a level can be reached, but not always how many times. Also compare interior maxima, included endpoints, monotonic branches and end behaviour
4. 4Treat tangency separately A horizontal line touching an extremum still gives one intersection at that input. A double algebraic root is not two distinct points, and need not cross the line
5. 5Complete written work is ungraded Write the full parameter classification, proof and labelled comparison sketches

Family member and level

\(\displaystyle (x^2-2)^2=4\)

For a=2, which description counts the distinct intersections correctly?

Worked model: compare a central maximum and minima

For a≥0, solve (x²−a)²=4 by writing x²=a+2 or x²=a−2. The first is always positive. The second is negative for 0≤a<2, zero at a=2, and positive for a>2. Thus the counts are 2, 3 and 4 respectively. For a>0, the minima have height 0 at ±√a and the central maximum has height a². It reaches level 4 exactly at a=2. At a=0 the central point is instead a flat minimum, but the count remains 2. Both tails rise

1. 1Count distinct allowed inputs After substituting u = x², keep only u ≥ 0. A positive u gives two real inputs, zero gives one, and a repeated u is not counted twice. Apply any original domain restriction afterwards \(\displaystyle u>0\Rightarrow x=\pm\sqrt u;\quad u=0\Rightarrow x=0\)
2. 2Compare every relevant height A minimum says whether a level can be reached, but not always how many times. Also compare interior maxima, included endpoints, monotonic branches and end behaviour
3. 3Treat tangency separately A horizontal line touching an extremum still gives one intersection at that input. A double algebraic root is not two distinct points, and need not cross the line

Your complete minimum and intersection-count investigation

For every t>0, derive all stationary inputs and classify them, treating t=1/2 separately. Find the global minimum value and where it is attained. Then give the complete number of distinct intersections with y=1 as t varies, including every threshold and equality case. Explain the role of the central height and why the minimum alone is insufficient. Independently check your counts using u=x² and retain only u≥0. Draw labelled comparison sketches for one parameter in each different count case; do not use a list of numerical examples as the proof

1. 1Complete written work is ungraded Write the full parameter classification, proof and labelled comparison sketches

Family and conditions

\(\displaystyle q_t(x)=(x^2-t)^2+x^2=1,\quad x\in\mathbb R,\quad t=9/8\)

After completing the classification, how many distinct intersections occur when t=9/8?

Review every parameter interval and boundary

The derivative is 2x(2x²−2t+1). For 0<t≤1/2, zero is the only stationary input and gives the global minimum t²; at t=1/2 this minimum is flat. For t>1/2, zero is a local maximum of height t², while ±√(t−1/2) are global minima of height t−1/4. Both tails rise. Relative to y=1: there are 2 intersections for 0<t<1; 3 at t=1; 4 for 1<t<5/4; 2 at t=5/4; and none for t>5/4. At t=1 the points are (−1,1),(0,1),(1,1). At t=5/4 the two minima touch the line at x=±√3/2. Compare the heights and monotonic branches, rather than doubling a repeated root

1. 1Complete written work is ungraded Write the full parameter classification, proof and labelled comparison sketches

Check the count without relying on the drawing

For u=x², the discriminant is 5−4t. If 0<t<1, the quadratic product t²−1 is negative, so exactly one u is positive and one is negative. At t=1 the squared inputs are 0 and 1. For 1<t<5/4, the discriminant is positive and both the product and sum 2t−1 are positive, so two distinct positive u values give four inputs. At t=5/4 there is one repeated positive u=3/4, giving two distinct inputs. For larger t there are no real u values. This independently accounts for all positive t.

1. 1Count distinct allowed inputs After substituting u = x², keep only u ≥ 0. A positive u gives two real inputs, zero gives one, and a repeated u is not counted twice. Apply any original domain restriction afterwards \(\displaystyle u>0\Rightarrow x=\pm\sqrt u;\quad u=0\Rightarrow x=0\)
2. 2Treat tangency separately A horizontal line touching an extremum still gives one intersection at that input. A double algebraic root is not two distinct points, and need not cross the line

Transfer the argument to a restricted domain

Repeat the complete intersection-count classification for the same family restricted to x>0. Give exact coordinates when t=1 and t=5/4, and explain why simply halving every previous count fails at t=1. Also state the global minimum or unattained lower bound on x>0 for 0<t≤1/2. Use the original restriction, not just the quadratic discriminant

1. 1Complete written work is ungraded Write the full parameter classification, proof and labelled comparison sketches

Family and conditions

\(\displaystyle (x^2-1)^2+x^2=1,\quad x>0\)

On the restricted domain x>0, how many intersections occur when t=1?

Review excluded endpoints and distinct points

On x>0, the counts are 1 for 0<t≤1, 2 for 1<t<5/4, 1 at t=5/4, and 0 for t>5/4. At t=1 only (1,1) remains; at t=5/4 only (√3/2,1) remains. The earlier three-point case included zero, so halving it was not valid. For 0<t≤1/2, the lower bound is t² as x approaches zero from above, but it is not attained: there is no global minimum. For t>1/2 the positive minimum input √(t−1/2) is allowed

1. 1Complete written work is ungraded Write the full parameter classification, proof and labelled comparison sketches

Derive the intercept function before optimising it

For a contact input a, form the tangent with the original height and derivative. Its y-intercept is B(a)=f(a)−af′(a); differentiating cancels the two f′ terms, leaving B′(a)=−af″(a). If f′(a)≠0, its unique x-intercept is a−f(a)/f′(a). A horizontal tangent needs separate handling: it has no x-intercept unless its height is zero, in which case the whole tangent is the x-axis. Intercept optimisation concerns the parameter a, not the running coordinate x on the line

1. 1Find the point At x = a, the point on the curve has coordinates (a, f(a)) \(\displaystyle P=\left(a,f(a)\right)\)
2. 2Find the tangent gradient Differentiate the function and substitute the x-coordinate of the point \(\displaystyle m_{\mathrm{t}}=f'(a)\)
3. 3Use point-gradient form Substitute the point and the required gradient into the equation of a straight line \(\displaystyle y-y_0=m\left(x-x_0\right)\)

Your tangent, intercept-extrema and shared-intercept synthesis

For f(x)=sin x−1/2 and contact inputs a∈[0,2π], derive the full tangent Tₐ(x) and its y-intercept B(a). Determine the exact global minimum and maximum possible y-intercepts, checking all stationary and boundary candidates. Next find every a for which Tₐ has a unique x-intercept equal to the curve intercept R=(π/6,0). Keep any exact contact value; report other contact inputs to three decimal places with independent brackets. Explain why horizontal tangents are excluded and prove your list is complete. Keep the derivation, radian-mode numerical record, signs and a drawing of the two qualifying tangents through R.

1. 1Complete work is ungraded The full written work, numerical record and drawing are ungraded

Curve, tangent and conditions

\(\displaystyle f(x)=\sin x-\tfrac12,\quad T_a(x)=\cos a(x-a)+\sin a-\tfrac12,\quad T_a(\pi/6)=0,\quad\cos a\ne0,\quad0\le a\le2\pi\)

After proving your complete list, how many distinct contact inputs give a tangent with the required unique x-intercept?

Review the line and all extrema candidates

The tangent, its y-intercept and the intercept derivative are shown above. It increases on (0,π) and decreases on (π,2π). The candidate values are B(0)=−1/2, B(π)=π−1/2 and B(2π)=−2π−1/2. Therefore the global maximum is π−1/2 at a=π, and the global minimum is −2π−1/2 at a=2π. The shared-intercept contacts are a=π/6 and a≈4.327885227462; the second reports as 4.328. Neither has zero cosine. At a=π/2 and 3π/2, the tangents are y=1/2 and y=−3/2, so they have no x-intercept

1. 1Complete work is ungraded Keep your derivation, radian-mode input/settings, unrounded iteration record, graph with labelled contacts and tangent intercepts, independent signs and explanation. The full written/technology work and drawing are ungraded

Prove that a numerical solver has not missed another contact

Let r=π/6 and H(a)=sin a−1/2−(a−r)cos a. Then H′=(a−r)sin a. H decreases from H(0)=r−1/2>0 to H(r)=0, then increases to H(π)=π−r−1/2>0. It decreases strictly on (π,2π), and H(3π/2)=−3/2<0. Thus there is exactly one further zero in (π,3π/2), and none elsewhere. The exact zero at r does not change sign, so a sign-change scan alone could miss it. Independent evaluations give H(4.32788)>0 and H(4.32789)<0, which certify the second contact to three places. A valid computation may use bisection or safeguarded Newton for this second root

1. 1Complete work is ungraded Keep your derivation, radian-mode input/settings, unrounded iteration record, graph with labelled contacts and tangent intercepts, independent signs and explanation. The full written/technology work and drawing are ungraded

Certified enclosure

\(\displaystyle 4.32788<\alpha<4.32789\)

Use this independently certified enclosure to report the indicated input to exactly three decimal places