Sketch f(x) = -2x^2 * e^x from x = -7 to x = 0, identifying the critical points, inflection points and range.

Step-by-step solution

1. 1

   Read the sign and the endpoints first. Since $x^2\ge 0$ and $e^x>0$ always, the factor $-2$ forces

   $f(x)\le 0\quad\text{for all }x,\qquad f(x)=0\text{ only at }x=0$

   At the left endpoint $f(-7)=-98e^{-7}\approx -0.0894$, so the curve starts just below the axis — the exponential has already crushed the $x^2$ growth.

2. 2

   Differentiate with the product rule.

   $f^{\prime}(x)=-2\left(2xe^{x}+x^{2}e^{x}\right)=-2xe^{x}(x+2)$

   so the critical points are $x=0$ and $x=-2$; the factor $e^x$ never vanishes.

3. 3

   Determine the direction of travel on each piece. For $x<-2$ both $x$ and $x+2$ are negative, so their product is positive and $f^{\prime}<0$: the curve falls. For $-2<x<0$ the product is negative, so $f^{\prime}>0$: the curve rises. Hence

   $x=-2\ \text{is a local minimum},\qquad f(-2)=-2(4)e^{-2}=-\frac{8}{e^{2}}\approx -1.0827$

   and $x=0$ is the maximum on this interval, with $f(0)=0$.

4. 4

   Find the inflection points from the second derivative.

   $f^{\prime\prime}(x)=-2e^{x}\left(x^{2}+4x+2\right)$

   $x^{2}+4x+2=0\;\Longrightarrow\;x=-2\pm\sqrt2\approx -3.4142,\ -0.5858$

   with $f(-3.4142)\approx -0.7671$ and $f(-0.5858)\approx -0.3820$. The curve is concave up between them and concave down outside.

5. 5

   Assemble the sketch and state the range. On $[-7,0]$ the graph starts near $-0.089$, descends to the trough $\left(-2,\,-8e^{-2}\right)$, then climbs back to touch the origin from below, flattening as it arrives because $f^{\prime}(0)=0$. The range is therefore

   $\left[-\frac{8}{e^{2}},\,0\right]\approx[-1.0827,\,0]$

   A quick numeric check confirms the shape: $f(-7)=-0.0894$, $f(-3.4142)=-0.7671$, $f(-2)=-1.0827$, $f(-0.5858)=-0.3820$, $f(0)=0\;\checkmark$.