Is #f(x)=1-x-e^(-3x)/x# concave or convex at #x=4#?

1 Answer
Feb 5, 2018

Let's take some derivatives!

Explanation:

For #f(x) = 1 - x - e^(-3x)/x#, we have
#f'(x) = - 1 - (-3xe^(-3x)-e^(-3x))/x^2#
This simplifies (sort of) to
#f'(x) = - 1 + e^(-3x)(3x+1)/x^2#
Therefore
#f''(x) = e^(-3x)(-3x-2)/x^3-3e^(-3x)(3x+1)/x^2#
# = e^(-3x)((-3x-2)/x^3-3(3x+1)/x^2)#
# = e^(-3x)((-3x-2)/x^3+(-9x-3)/x^2)#
# = e^(-3x)((-3x-2)/x^3+(-9x^2-3x)/x^3)#
# = e^(-3x)((-9x^2-6x-2)/x^3)#

Now let x = 4.

#f''(4) = e^(-12)((-9(16)^2-6(4)-2)/4^3)#

Observe that the exponential is always positive. The numerator of the fraction is negative for all positive values of x. The denominator is positive for positive values of x.

Therefore #f''(4) < 0#.

Draw your conclusion about concavity.