What are the points of inflection, if any, of #f(x)=2x^4-e^(8x#?

1 Answer
Apr 24, 2018

See below

Explanation:

First step is finding the second derivative of the function

#f(x)=2x^4-e^(8x)#

#f'(x)=8x^3-8e^(8x)#

#f''(x)=24x^2-64e^(8x)#

Then we must find a value of x where:

#f''(x)=0#

(I used a calculator to solve this)

#x=-0.3706965#

So at the given #x#-value, the second derivative is 0. However, in order for it to be a point of inflection, there must be a sign change around this #x# value.

Hence we can plug values into the function and see what happens:

#f(-1)=24-64e^(-8)# definetly positive as #64e^(-8)# is very small.

#f(1)=24-64e^(8)# definetly negative as #64e^8# is very big.

So there is a sign change around #x=-0.3706965#, so it is therefore an inflection point.