# What is the slope of f(x)=-e^(x-3x^3)  at x=-2?

Feb 5, 2016

The slope of $f \left(x\right)$ at $x = - 2$ is extremely high, $\approx 1.25 \cdot {10}^{11}$.

#### Explanation:

To calculate the slope of $f \left(x\right)$, you need to compute the derivative of $f \left(x\right)$ first.

Use the chain rule for that:

$f \left(x\right) = - {e}^{x - 3 {x}^{3}} = - {e}^{u} \text{ }$ where $\text{ } u = x - 3 {x}^{3}$

Thus, the derivative of $f \left(x\right)$ is:

$f ' \left(x\right) = \left[- {e}^{u}\right] ' \cdot u ' = - {e}^{u} \cdot \left(1 - 9 {x}^{2}\right) = - {e}^{x - 3 {x}^{3}} \left(1 - 9 {x}^{2}\right)$

$= \left(9 {x}^{2} - 1\right) {e}^{x - 3 {x}^{3}}$

Now, you will find the slope of $f \left(x\right)$ at $x = - 2$ if you evaluate $f ' \left(- 2\right)$:

$f ' \left(- 2\right) = \left(9 {\left(- 2\right)}^{2} - 1\right) {e}^{- 2 - 3 \cdot {\left(- 2\right)}^{3}} = 35 {e}^{22}$

$\approx 125471949614.6 \approx 1.25 \cdot {10}^{11}$

This is extremely steep. Let's plot the function to see if this number might make sense:

graph{- e^(x- 3x^3) [-14.03, 18.01, -13.49, 2.53]}

The function looks very steep indeed for $x < - 1$.