Graphs of f, f' & f'' (College Board AP® Calculus AB): Revision Note

Make sure you refer explicitly to the relevant function in your answer. It can be confusing if you say "it is continuous" because the reader will not know what you mean by "it". It is much better to use the name of the function, such as, "$f\text{'}$ is continuous".

The graph of $f^{\text{'}}$, the derivative of the function $f$, is shown above. It has a local maximum at $x=-2$ and a local minimum at $\left(0,0\right)$. Which of the following statements must be true?

i. $f$ has a critical point at $x=0$.

ii. The graph of $f$ has a point of inflection at $x=-2$.

iii. The graph of $f$ is concave up for $-7<x<1$.

iv. The function $f$ is increasing for $-7<x<1$.

(A) i and ii only (B) i, ii, and iii only (C) i, ii, and iv only (D) i, ii, iii, and iv

Answer:

Let's look at each statement in turn to determine whether it is true or not

i. $f$ has a critical point at $x=0$.

• This is true as $f^{\text{'}}\left(x\right)=0$ at $x=0$

• therefore it is a critical point

ii. The graph of $f$ has a point of inflection at $x=-2$.

• A point of inflection must have $f^{\text{'}\text{'}}\left(x\right)=0$, so we must consider the slope of the graph of $f^{\text{'}}\left(x\right)$

  • A tangent drawn at $x=-2$ on $f^{\text{'}}\left(x\right)$ would be horizontal

  • therefore $f^{\text{'}\text{'}}\left(x\right)=0$, so it could be a point of inflection

• We still need to check whether the second derivative changes sign

  • To the left of $x=-2$, the graph of $f^{\text{'}}$ is increasing, so $f^{\text{'}\text{'}}$ will be positive

  • To the right of $x=-2$, the graph of $f^{\text{'}}$ is decreasing, so $f^{\text{'}\text{'}}$ will be negative

  • The sign of the second derivative changes at $x=-2$, so it is a point of inflection

• Therefore statement ii is true

iii. The graph of $f$ is concave up for $-7<x<1$.

• Concave up is when $f^{\text{'}\text{'}}\left(x\right)>0$, so we need to consider whether or not the slope of $f^{\text{'}}\left(x\right)$ is always positive

• The graph of $f^{\text{'}}\left(x\right)$ slopes downwards between $x=-2$ and $x=0$

  • therefore $f^{\text{'}}$ is decreasing on that interval; $f^{\text{'}\text{'}}$ is negative

• Therefore statement iii is false.

iv. The function $f$ is increasing for $-7<x<1$.

• For a function to be increasing, $f^{\text{'}}\left(x\right)$ must be non-negative

• The graph of $f^{\text{'}}\left(x\right)$ is above or on the $x$-axis between -7 and 1, therefore $f\left(x\right)$ is always increasing in this interval

• Therefore statement iv is correct

So statements i, ii, and iv are true

Option (C)

Let $f$ be the function is defined by $f\left(x\right)=x{\left(x-2\right)}^2\left(x+3\right)$.

Sketch the graph of $y=f\left(x\right)$, labeling the coordinates of any axes intercepts and critical points, and stating the $x$ values of any points of inflection.

Answer:

By inspecting the function, it can be seen that it is a positive quartic (i.e. highest power of $x$ is 4)

This means overall it will have a w-shape, and will tend toward $+\infty$ as $x$ tends to $\pm \infty$

The roots can also be found easily, by setting $f\left(x\right)=0$

$x{\left(x-2\right)}^2\left(x+3\right)=0$

Roots at $x=0$ and $x=-3$, and a repeated root at $x=2$

Find any critical points where the first derivative is zero

Expand the equation first to make this easier

$f\left(x\right)=x^4-x^3-8x^2+12x$

$f^{\text{'}}\left(x\right)=4x^3-3x^2-16x+12=0$

You could solve this on your calculator, or by factoring

$f^{\text{'}}\left(x\right)=\left(x-2\right)\left(4x^2+5x-6\right)=\left(x-2\right)\left(4x-3\right)\left(x+2\right)=0$

Critical points at $x=2,x=\frac{3}{4},x=-2$

Put these values into $f\left(x\right)$ to find the $y$-coordinates

Critical points at (2, 0), (-2, -32), and $\left(\frac{3}{4},\frac{1125}{256}\right)$

This should be enough information to sketch the shape of the graph and its critical points, but it can be useful to check the concavity of each critical point

Use the second derivative to do this

$f^{\text{'}\text{'}}\left(x\right)=12x^2-6x-16$

$f^{\text{'}\text{'}}\left(2\right)=20$, positive, so u-shaped at $x=2$ (a minimum)

$f^{\text{'}\text{'}}\left(\frac{3}{4}\right)=-\frac{55}{4}$, negative, so n-shaped at $x=\frac{3}{4}$ (a maximum)

$f^{\text{'}\text{'}}\left(-2\right)=44$, positive, so u-shaped at $x=-2$ (a minimum)

Check for any points of inflection, where the second derivative is equal to zero

All points of inflection have $f\text{'}\text{'}\left(x\right)=0$, but some critical points will also have $f\text{'}\text{'}\left(x\right)=0$. However we have already found all the critical points, so any remaining points where  $f\text{'}\text{'}\left(x\right)=0$ must be a point of inflection

$f^{\text{'}\text{'}}\left(x\right)=12x^2-6x-16=0$

$x=\frac{3\pm \sqrt{201}}{12}$

$x=1.431453907...$ or $x=-0.9314539066...$

These points of inflection are where the graph changes concavity, so will be in between the local minimums and maximums

Sketch the graph, labeling all the points as asked in the question