a) For what value of x does f reach its absolute maximum? justify your answers
b) Find the x-coordinates of all points of inflection of f. Justify your answers.f: [-4, 4] ¡ú R, f(x) = (x2+1)e^(-x)
f'(x) = 2xe^(-x) - (x2 + 1)e^(-x) = -(x2 - 2x + 1)e^(-x) = -(x+1)2e^(-x)
f''(x) = -2(x+1)e^(-x) + (x+1)2e^(-x)
Now, f will attain its maximum on [-4, 4] either at a critical point, or at some point on the boundary of [-4, 4]. The boundary points of [-4, 4] are of course, -4 and 4. Now, let us find the critical points of f -- these are the points where f'(x) = 0, which means:
-(x+1)2e^(-x) = 0
(x+1)2 = 0 (dividing by -e^(-x) is justified, since -e^(-x) is never 0)
x+1 = 0
x = -1
So f will reach its maximum either at -4, -1, or 4. Evaluating f at these three points yields:
f(-4) = ((-4)2 + 1)e^(-(-4)) = 17e^4
f(-1) = ((-1)2 + 1)e^(-(-1)) = 2e
f(4) = (42 + 1)e^(-4) = 17e^(-4)
Clearly, the maximum value occurs at x = -4. Another way of seeing this is that f'(x) is always less than or equal to zero, so f is a monotone decreasing function, so the maximum will occur at the leftmost point of the domain.
To find the points of inflection, these are the points where f''(x) = 0, which means:
-2(x+1)e^(-x) + (x+1)2e^(-x) = 0
(x+1)2 - 2(x+1) = 0
(x+1 - 2)(x+1) = 0
(x-1)(x+1) = 0
x = 1 or x = -1
So the points of inflection are 1 and -1. And we are done.