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

[karush]

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 ?

 

[skeeter]

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]

 

[skeeter]

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

 

[karush]

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]
\addplot[
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}
$$e^{(x^2-1)^2}-2$$

 

[HOI]

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 ?