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

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Answer 1 · 2016-02-05T15:13:19

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

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

Use the chain rule for that:

$f (x) = - e^{x - 3 x^{3}} = - e^{u}$ $f(x) = - e^(x- 3x^3) = -e^u" "$ where $u = x - 3 x^{3}$ $" "u = x - 3x^3$

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

$f'(x) = [-e^u]' * u' = -e^u * (1 - 9 x^2) = -e^(x - 3x^3) (1 - 9x^2)$

$= (9x^2 - 1) e^(x - 3x^3)$

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

$f'(-2) = (9 (-2)^2 -1) e^(-2 - 3 * (-2)^3) = 35 e^(22)$

$~~125471949614.6 ~~1.25 * 10^11$

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 $x < -1$ .

What is the slope of f(x)=-e^(x-3x^3) f(x)=−ex−3x3f(x)=-e^(x-3x^3) at x=-2x=−2x=-2?