by the product rule ...
f '(x) = (x^2 + 1)(-1)e^(-x) + 2x e^-x
= e^-x (-x^2 - 1 + 2x)
for a max or min, f '(x) = 0
e^-x (-x^2 - 1 + 2x)
e^-x = 0 ---> not possible
or
x^2 -2x + 1 = 0
(x-1)^2 =
x = 1
f(1) = (1+1)e^-1= 2/e = appr .736
check the ends since we are given a domain:
f(-4) = 17*e^4 = appr 928
f(4) = 17/e^4 = appr .311
So x = -4 yields the maximum
while x=1 would yield a local maximum
points of inflection f ''(x) = 0
f ''(x) = (x-1)^2 (-e^-x) + 2(x-1)e^-x
= e^-x(-(x-1)^2 + 2(x-1)
= e^-x (-x^2 + 4x -3)
= 0
x^2 - 4x + 3 = 0
(x-1)(x-3) = 0
points of inflection at x = 1 and x = 3
observe (1, 2/e) is both a turning point and a point of inflection
Let f be the function defined by f(x) = (x^2 + 1)e^-x for -4≤x≤4.
a. For what value of x does f reach its absolute maximum? Justify your answer. b. Find the x-coordinates of all points of inflection of f. Justify your answer.
1 answer
1 answer