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

Question

William Abdalla · Accepted Answer

See below

Finding the function's second derivative is the first step. #f(x)=2x^4-e^(8x)# #f'(x)=8x^3-8e^(8x)# #f''(x)=24x^2-64e^(8x)# Next, we need to determine x's value where: #f''(x)=0# (I solved this using a calculator.) #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. Thus, we can enter values into the function to observe the following outcome: #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.

Isaiah Allen · Answer

undefined To find the points of inflection of $ f(x) = 2x^4 - e^{8x} $, we first need to find the second derivative and then determine where it equals zero.

$$ f(x) = 2x^4 - e^{8x} $$
$$ f'(x) = 8x^3 - 8e^{8x} $$
$$ f''(x) = 24x^2 - 64e^{8x} $$

Setting $ f''(x) $ equal to zero and solving for $ x $:

$$ 24x^2 - 64e^{8x} = 0 $$
$$ 24x^2 = 64e^{8x} $$
$$ x^2 = \frac{64}{24}e^{8x} $$
$$ x^2 = \frac{8}{3}e^{8x} $$

This equation cannot be solved algebraically. Points of inflection occur where the concavity of the function changes, which happens when the second derivative crosses the x-axis. To find approximate points of inflection, numerical methods or graphing software can be used.