# 3.3.2 AP Calculus Exam interval from f'(x)

• MHB
• karush
In summary, the conversation discusses how to find the solution to a problem involving the function $f(x) = e^{(x^2-1)^2}-2$. The participants suggest graphing the first and second derivatives of the function and looking for intervals where the slope of the first derivative is less than the slope of the second derivative. They also mention that $f'(0)$ does not equal any of the given numbers and clarify that the statement "concave down" refers to the negative value of the second derivative.
karush
Gold Member
MHB

screen shot to avoid typos

OK the key said it was D

I surfed for about half hour trying to find a solution to this but $f'(0)$ doesn't equal any of these numbers

$e^0=\pm 1$ from the $e^{(x^2-1)^2}$

kinda ?

this is a calculator active problem.
graph f’(x) in the given interval and look for the intervals where the slope of f’(x) = f’’(x) < 0

$(-1.5,-1) \cup (0,1)$

[DESMOS]{"version":7,"graph":{"viewport":{"xmin":-3,"ymin":-12.12442693138332,"xmax":3,"ymax":12.12442693138332}},"randomSeed":"f9c61ea916ed9ae4b364c88ac8d063a8","expressions":{"list":[{"type":"expression","id":"1","color":"#c74440","latex":"y=\\exp\\left(\\left(x^{2}-1\\right)^{2}\\right)-2"}]}}[/DESMOS]

this can also be done algebraically ...

$f’’(x) = e^{(x^2-1)^2} \cdot 4x(x^2-1)$

$f’’(x) =0$ at $x=0$ and $x= \pm 1$

from here one can determine the sign of f’’(x) in the four intervals bounded by the zeros

my feeble attempt at tikz

\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[xmin=-2, xmax=2, ymin=-3, ymax=3, axis lines=middle, ticks=none]
draw = blue, smooth, ultra thick,
domain=-1.5:1.5,
] {exp((x^2-1)^2)-2}
foreach \x in {-1.5,-1,1,1.5} { (axis cs:{\x},0) node[below left] {\x} };
\end{axis}
\end{tikzpicture}
$$e^{(x^2-1)^2}-2$$

Last edited:
karush said:
View attachment 10721
screen shot to avoid typos

OK the key said it was D

I surfed for about half hour trying to find a solution to this but $f'(0)$ doesn't equal any of these numbers
Why should it be? Saying that the graph is ''concave down" means that the second derivative is negative and that changes where the second derivative, not the first, is 0.

$e^0=\pm 1$ from the $e^{(x^2-1)^2}$

kinda ?

## 1. What is the purpose of finding the interval from f'(x) on the AP Calculus Exam?

The interval from f'(x) on the AP Calculus Exam is used to determine the critical points of a function, which are points where the derivative is equal to zero or undefined. This information is important for analyzing the behavior of a function and solving optimization problems.

## 2. How do you find the interval from f'(x) on the AP Calculus Exam?

To find the interval from f'(x) on the AP Calculus Exam, you first need to find the derivative of the given function. Then, set the derivative equal to zero and solve for x. The resulting values of x will be the critical points. Finally, use the critical points to create intervals on a number line to determine the intervals where the derivative is positive or negative.

## 3. Can the interval from f'(x) on the AP Calculus Exam have more than two intervals?

Yes, the interval from f'(x) on the AP Calculus Exam can have more than two intervals. This can occur when there are multiple critical points or when the function has a discontinuity at a critical point.

## 4. How is the interval from f'(x) related to the concavity of a function?

The interval from f'(x) is related to the concavity of a function because it helps determine where the function is increasing or decreasing. If the derivative is positive on an interval, the function is increasing and the concavity is positive. If the derivative is negative on an interval, the function is decreasing and the concavity is negative.

## 5. Is it necessary to find the interval from f'(x) on the AP Calculus Exam if the function is already given in its graph form?

Yes, it is still necessary to find the interval from f'(x) on the AP Calculus Exam even if the function is given in its graph form. This is because the graph may not accurately show the exact values of the critical points, and finding the intervals from f'(x) will provide more precise information for analyzing the function.

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