1. Use the Mean Value Theorem to find an x-value where the instantaneous slope=average slope over the given interval:

Question

f(x)=(9-x^2)^(1/2) on [-3,2]
I have done three chapters of Calculus already and I never have had to check for extraneous roots. So, my question is, when exactly do I check for extraneous roots? (ex/ when I was looking for critical points and I root my answer, I didn't have to check for extraneous roots)

2. Can someone show me how e^(ln((e^5-1)/5))=(e^5-1)/5?

Answer 1

by definition, e^(ln x) = x

for the MVT,

f(-3) = 0
f(2) = √5
average rate is thus √5/5 or 1/√5

So, you want to find where the slope of f is 1/√5

f' = -x/√(9-x^2)

so, find x where f' = 1/√5

-x/√(9-x^2) = 1/√5

square both sides. (This is where an extraneous root might appear, because both (√a)^2 = a and (-√a)^2 = a)

x^2/(9-x^2) = 1/5
5x^2 = 9-x^2
6x^2 = 9
x = ±3/√6

Now you have to check which solution you want.

f'(3/√6) = -(3/√6)/√(9 - 9/6)
= -3/√6 / √(45/6)
= -3/√45
= -1/√5

And of course, f'(-3/√6) = +1/√5

See the graphs at

http://www.wolframalpha.com/input/?i=plot+y%3D%E2%88%9A%289-x^2%29%2C+y%3D1%2F%E2%88%9A5%28x%2B3%29%2C+y+%3D+1%2F%E2%88%9A5+%28x+%2B+3%2F%E2%88%9A6%29+%2B+%E2%88%9A%2845%2F6%29+where+-3+%3C%3Dx+%3C%3D+2