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

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Answer 1 · 2016-02-05T15:13:19

The slope of #f(x)# at #x = -2# is extremely high, #~~1.25 * 10^11#.

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

Use the chain rule for that:

#f(x) = - e^(x- 3x^3) = -e^u" "# where #" "u = x - 3x^3#

Thus, the derivative of #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#.