Mean Value Theorem answer help

[karisrou]

1. If c is the value defined by the mean value theorem, then for f(x) = e^x - x^2 on [0,1], c=

I found the two end points as [0,1] and [1,e-1], so the average slope is .71828...

is that the answer then?

 

[tnutty]

nope, what's the formula for the MVT?

 

[karisrou]

f(b)-f(a) / b-a

so (e-1) - (1-1) / (1 - 0)

Which gives 1.718...

 

[n!kofeyn]

1. If c is the value defined by the mean value theorem, then for f(x) = e^x - x^2 on [0,1], c=?

How does your book state the mean value theorem? This is important. My book states that if f is continuous on [a,b] and f is differentiable on (a,b), then there exists a number c in (a,b) such that
[tex] f'(c) = \frac{f(b) - f(a)}{b-a} [/itex]

You found the quotient
$$\frac{f(1) - f(0)}{1-0} = e-2$$
(Don't put it in decimal form.)

Now if your book states the mean value theorem this way (which it most likely does), then you need to find the c in (a,b) such that f'(c)=e-2, which means you have an incorrect answer.