Mean Value Theorem answer help

karisrou
Messages
15
Reaction score
0
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?
 
Physics news on Phys.org
nope, what's the formula for the MVT?
 
f(b)-f(a) / b-a

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

Which gives 1.718...
 
karisrou said:
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
f&#039;(c) = \frac{f(b) - f(a)}{b-a} [/itex]<br /> <br /> You found the quotient<br /> \frac{f(1) - f(0)}{1-0} = e-2<br /> (Don&#039;t put it in decimal form.)<br /> <br /> 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&#039;(c)=e-2, which means you have an incorrect answer.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Replies
2
Views
1K
Replies
12
Views
1K
Replies
2
Views
1K
Replies
5
Views
2K
Replies
3
Views
1K
Replies
6
Views
1K
Replies
11
Views
2K
Back
Top