Second derivatives and anti-derivative graphs

Use first derivatives to interpret graphs

The first derivative gives tangent gradient. Its sign controls increasing/decreasing behaviour, and its sign changes classify stationary points. Differentiate f′ to study concavity, and use F′=g to recover anti-derivative graph features. Apply polynomial and chain rules where needed

Second derivative notation and meaning

For y=f(x), the second derivative f″(x)=d²y/dx² means differentiate dy/dx with respect to x again. It is not[f′(x)]² and not a ratio of ordinary squares. If y has units metres and x seconds, f′ has units m/s and f″ has units m/s². The sign of f″ describes how the gradient changes, not the sign of the height

Differentiate a polynomial twice

For f(x)=2x⁴−3x³+x²+5x−7, the first derivative is8x³−9x²+2x+5. Differentiating that entire expression gives f″(x)=24x²−18x+2. The constant term vanishes immediately; the original linear term vanishes on the second differentiation

Give the coefficient of x in f″(x)=24x²−18x+2.

Apply the chain rule again

For f(x)=−3sin(2x+1)+4, f′(x)=−6cos(2x+1), then f″(x)=12sin(2x+1). Each differentiation contributes the inner factor2. For f(x)=2e^(3x−1), the derivatives are6e^(3x−1) and18e^(3x−1). A sine or cosine returns with a minus sign after two derivatives; an exponential does not

Give K when the second derivative of−3sin(2x+1)+4 is Ksin(2x+1)

Concavity describes changing gradient

On an interval, f″>0 means f′ increases and the graph is concave up; f″<0 means f′ decreases and the graph is concave down. A graph can be falling while its gradient becomes less negative. Thus increasing/decreasing and concave up/down are separate descriptions. Isolated zeros of f″ do not by themselves split a concavity interval

For the plotted cubic, what happens immediately to the right of x=1?

A nonstationary inflection

For f(x)=(x−1)³−2(x−1)+1, f″(x)=6(x−1). It changes from negative to positive at x=1. The function is defined and continuous there, so(1,1) is an inflection. Its tangent gradient is f′(1)=−2, so it is not a stationary point. Its tangent is y−1=−2(x−1)

For f(x)=(x−1)³−2(x−1)+1, give the tangent gradient at its inflection point (1,1)

A zero second derivative is a candidate

For x³ and x⁴, both first and second derivatives are zero at0. But(x³)″=6x changes sign, while(x⁴)″=12x² is positive on both sides. The cubic has a stationary inflection; the quartic has a flat local minimum. Always examine concavity on both sides of a candidate

Which plotted curve has an inflection at0?

The second-derivative test has a condition

At a stationary point of a sufficiently differentiable function, f″>0 establishes a local minimum and f″<0 a local maximum. If f″=0, this test is inconclusive: use signs of f′ for turning behaviour and signs of f″ for concavity. A lack of sign change of f′ alone is not a general proof of inflection

At a stationary point, f″=0. What can the second-derivative test alone establish?

Find both inflections of a quartic

Let f(x)=−2[(x−1)⁴−6(x−1)²]+3. Then f″(x)=−24[(x−1)²−1]. Its roots are0 and2; the signs are negative, positive, negative on the three successive intervals. Substitution into f gives13 at both roots. Thus(0,13) and(2,13) are inflections, with concavity changing at each

For f(x)=−2[(x−1)⁴−6(x−1)²]+3, give the common ordinate of its inflections at x=0 and x=2.

Read three linked graphs

For f(x)=x³−3x, f′=3x²−3 and f″=6x. The zeros±1 of the quadratic locate the cubic stationary points. The zero0 of the line locates the quadratic turning point and the cubic inflection. A positive line value means the quadratic is increasing and the cubic is concave up. Derivative graphs are not generally vertically shifted copies of the original

For f(x)=x³−3x, f′(x)=3x²−3 and f″(x)=6x, give the x-coordinate of the cubic inflection

Reverse the graph relationship

An anti-derivative F of g satisfies F′=g. Where g is positive, F increases; where g is negative, F decreases. A zero of g gives a stationary point of F, but its nature depends on the sign change of g. Since F″=g′, increasing g corresponds to concave-up F, and decreasing g to concave-down F. Heights of g are slopes of F.

Read anti-derivative features from a quadratic

For g(x)=2[(x+1)²−4], its zeros are−3 and1. Its signs are positive, negative, positive, so every anti-derivative F has a maximum at−3 and minimum at1. Since g′=4(x+1), F is concave down for x<−1 and up for x>−1; its inflection occurs at−1. No vertical position of F is determined yet

If F′(x)=g(x)=2[(x+1)²−4], give the inflection x-coordinate of F.

A graph family, then an initial value

For g(x)=2(x−1)³−3(x−1), verify by differentiation that F(x)=(x−1)⁴/2−3(x−1)²/2+C is an anti-derivative. The condition F(1)=4 fixes C=4. Before that condition, every real C gives a vertical translate with the same derivative. This task uses recognition and verification; later sections develop further integration techniques

Give the coefficient of (x−1)² in F(x)=(x−1)⁴/2−3(x−1)²/2+C.

What stays the same across the plotted anti-derivative family?

A hole is not an inflection

For f(x)=1/x, f″(x)=2/x³ changes sign across0, but the function is undefined there. There is no point on the graph at0, hence no inflection there. Analyse each domain interval and require an actual continuous graph point when testing inflections

A second derivative need not exist at an inflection

For f(x)=x|x|, the left rule is−x² and the right rule is x². The first derivative is2|x|, including f′(0)=0 by the difference quotient. Concavity is down on x<0 and up on x>0. The origin is a stationary inflection even though the left and right derivatives of f′ are−2 and2, so f″(0) does not exist

Justify the test boundaries, ungraded

Explain why f″=0 is neither a sufficient inflection test nor a test for a particular stationary nature. Contrast x³,x⁴,−x⁴,1/x and x|x| at0. State the role of the original domain and which derivative exists in each case

Model: use separate properties

The cubic has a stationary inflection; x⁴ has a flat minimum and−x⁴ a flat maximum. All three have first and second derivatives0 at0. The reciprocal has no graph point at0, so its cross-hole sign change gives no inflection. The function x|x| has a continuous graph, derivative0 and a concavity change at0, although its second derivative fails there. These examples separate a candidate equation, a stationary classification and a concavity change

Justify the constant difference, ungraded

Explain why two anti-derivatives of the same g differ by a constant on an interval. Then give two anti-derivatives of g=0 on R without0 whose difference is not one global constant. State why this does not contradict the interval result

Model: connected domain matters

For H=F1−F2 on an interval, H′=g−g=0. The mean value theorem on any closed subinterval gives H(b)−H(a)=H′(c)(b−a)=0, so H is constant. On R without0, take F1=0 everywhere, and F2=1 for x<0 and2 for x>0. Both derivatives are0 on their stated domain, but their difference has a different constant on each component. There is no interval in the domain crossing the hole

Construct a complete paired sketch, ungraded

For g(x)=x²−1, sketch g and two anti-derivatives satisfying F′=g with F(0)=0 and F(0)=2. Label stationary and inflection points with exact coordinates, increasing/decreasing and concavity intervals, and the vertical separation. Explain which facts are read directly from g and which need the initial value

Model: link heights, slopes and vertical shifts

The curves are F0=x³/3−x and F2=F0+2. For F0 the maximum is(−1,2/3), the minimum(1,−2/3) and inflection(0,0). For F2 the maximum is(−1,8/3), the minimum(1,4/3) and the inflection(0,2). Both increase on(−∞,−1) and(1,∞), decrease on(−1,1), are concave down for x<0 and up for x>0. The graph of g supplies these x-locations and behaviours; the initial value fixes the height. A complete learner sketch remains ungraded

Investigate second differences with technology, ungraded

For f(x)=x⁴−6x², compare[f(1+h)−2f(1)+f(1−h)]/h² with the exact f″(1). Try h=0.1 and0.01. Check signs either side of1 as well as the value at1. Record your own calculations, graph or code; a reference calculation is not evidence of your execution

Model: numerical support with a stated limit

The exact second derivative is12x²−12, so f″(1)=0 and concavity changes at1. Expansion shows the central second difference at1 equals2h² exactly. It gives0.02 and0.0002 for the stated increments, tending to0. Finite differences support the exact analysis; floating-point cancellation for extremely small increments can obscure it. The zero alone would not establish the sign change

Exact values and complete graph explanations

Find second derivatives, concavity signs, inflections, ordered graph labels and polynomial anti-derivative families in the stated forms