Expert: Ahmed Salami - 11/29/2008

Question
Given the function f(x)=e^-x^2

(a). Find the derivative of f
(b). Find the critical values
(c). Draw a sign diagram that shows the critical values and where the derivative is positive and negative.
(d). Find the relative maxima and relative minima (if any).
(e). Find the intervals where the function is increasing and where it is decreasing.
(f). Find the absolute minimum and absolute maximum on the interval [-2, 1].

***Ok i really need help with this. i think the derivative is -2xe^-x^2 but i didnt think it had any critical values....Can u please help me!?!?


Answer
Hi Melissa,
Sorry it took some time.
a)Yes, you're correct with the derivative.
The derivative f'(x) = -2xe^(-x^2)
b)Critical values occur on a curve when the derivative is zero (critical points) or in some cases undefined(singular points).
f'(x) = 0 when x = 0 (and undefined when x is +/- infinity)
c)From f'(x) = -2xe^(-x^2), it can be seen that the derivative is positive for negative values of x and vice versa.
d)At x = 0, f(x) = e^0 = 1
f''(x) = -2[x.(-2x).e^(-x^2) + e^(-x^2)]
      = -2[-2x^2.e^(-x^2) + e^(-x^2)]
      = -2e^(-x^2)[-2x^2 + 1]
      = 2e^(-x^2)[2x^2 - 1]
At x = 0
f''(x) = 2.e^0.[2(0) - 1]
      = 2.1.-1
      = -2
A negative number which indicates a relative maxima at (0,1)
e)A function is increasing where f'(x) > 0 and vice versa. The result coincides with (c) for this function.
f)From (d) we can see that there's only maximum value for the function at (0,1)

Regards.