posted by Isabella .
1. Use the Mean Value Theorem to find an x-value where the instantaneous slope=average slope over the given interval:
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?
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)
And of course, f'(-3/√6) = +1/√5
See the graphs at