Really stuck on Maths A2 Question?

Question

I have spent an hour trying to do this... please help.

Note: ^ means to the power of

f:x --> e^(1-x^2)

The minimum values of f(x), as x varies between 0 and 2, are m and M respectively.

Show that M-m = 2.67





So far i have differentiatied f:x to give me -2xe^1-x^2
dy/dx must = 0 fo a stationary points and so i know that when x = 0 there is a stationary point, but where is the other??

Forgot to add.. 0=

Answer 1

There is no other stationary point (df/dx = 0). so you know that the min/max are the ends of the range.
for x=0, f(x)= e^1 = e = 2.72
for x=2, f(x) is about 0.05
the change in value is 2.72-.05 = 2.67

Answer 2

The theorem says that the maximum and minimum values occur either at a critical point or at one of the endpoints of the interval. For a problem like this, you will always need to evaluate the function at both endpoints and consider those as possibilities for extrema.

Answer 3

up meh growth, lol joking the different table certain factor iw dyx it fairly is two^-4 . verify it in case you do no longer belive me yet i've got been given an a* in 'o' point maths. if i'm incorrect i will deliver you 50p on paypal lol. stable luck and verify out to no longer cheat back!