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

What should be reported at the excluded left endpoint?

Use the derivative to build the whole drawing

Keep the original function, derivative-sign chart and complete labelled drawing together. Whole drawings and explanations are ungraded; the small checked values assess individual features

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

Draw the curve from derivative evidence

For f(x)=x³−3x, the intercepts are x=−√3,0,√3. The derivative 3(x−1)(x+1) has signs +,−,+ on the intervals split at −1 and 1. Therefore the curve increases to the local maximum (−1,2), decreases to the local minimum (1,−2), then increases. Its first-derivative graph decreases up to x=0 and increases after x=0, identifying changing slope trend at (0,0), where the slope is −3. Domain and range are R; the left end falls and the right end rises. Draw one smooth curve consistent with the roots, heights and derivative signs

Rule

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

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

Your complete derivative-supported drawing

Draw the complete graph g(x)=(x−1)³−3(x−1) on R. Calculate and factor its first derivative; produce a sign chart, classify both stationary points using sign changes and calculate their heights using g. Label every intercept, the symmetry centre, domain/range and both ends. Use the derivative graph to describe the changing slope trend at the centre. Verify an extra exact point and connect a smooth curve consistent with every feature. Keep the sign chart beside your drawing

Review the sign chart and whole curve

g′(x)=3x(x−2), with signs +,−,+ on (−∞,0),(0,2),(2,∞). The local maximum is (0,2), the local minimum is (2,−2); these are heights of g, not of g′. The three simple roots are 1−√3,1,1+√3, all crossings. The centre is (1,0), where g′(1)=−3 and the slope trend changes from decreasing to increasing. Domain and range are R. The left end falls and the right end rises, so neither local extremum is global. An extra exact point is (3,2)

A quartic can touch both intercepts

For f(x)=(x²−1)², the first derivative is 4x(x−1)(x+1), with signs −,+,−,+ across the stationary inputs −1,0,1. The curve decreases, increases, decreases and increases in that order. Evaluate f to obtain minima (−1,0),(1,0) and the local maximum (0,1). Both roots are double and touch the axis. The graph is even, has domain R and range [0,∞), and both ends rise without bound. The minimum is global, but the local maximum is not global. Use these features to complete the W-shaped curve

Rule

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

Calculate the height of the worked quartic at the stationary input zero

Your complete derivative-supported drawing

Draw g(x)=(x²−4)² on the restricted domain [−3,3]. Find and factor g′, make a derivative-sign chart and evaluate every stationary input and both included endpoints. Label intercepts and touching behaviour, interior local extrema, global extrema, symmetry and the range. Mark both endpoints as filled and do not draw tails outside the specified domain. Then state what changes to the range and global maximum if both endpoints are removed, giving domain (−3,3)

Review the sign chart and whole curve

g′(x)=4x(x−2)(x+2), with signs −,+,−,+ on (−3,−2),(−2,0),(0,2),(2,3). The double roots (−2,0),(2,0) are global minima; the interior local maximum and y-intercept is (0,16). The filled endpoints (−3,25),(3,25) attain the global maximum 25. The graph is symmetric in the y-axis and its range is [0,25]. Restrict every branch to [−3,3]. With domain (−3,3), replace the endpoint dots by open ones: range [0,25), no global maximum, and supremum 25. The interior features do not change

A derivative zero need not produce a turn

Draw r(x)=(x−2)³ on R. Find r′, give its sign on both open intervals separated at 2, calculate the stationary point and every intercept, and label another point on each side. Explain why the horizontal tangent at the x-intercept is not a maximum or minimum. Compare the sign change at a simple stationary input of g(x)=(x−1)³−3(x−1) with this squared factor in r′

Use the derivative sign, not just its zero

r′(x)=3(x−2)² is positive on both (−∞,2) and (2,∞), and zero at 2. There is no sign change, so (2,0) is a stationary inflection, not an extremum. The cubic crosses its triple root while flattening. Its y-intercept is (0,−8); points (1,−1) and (3,1) show the two sides. Domain/range R, symmetry centre (2,0), left end falling and right end rising. The derivative graph decreases to zero then increases, so the slope trend changes at 2. A zero derivative alone did not justify either maximum or minimum