# Homework Help: Verification of Problems

1. Feb 9, 2005

Hello all

If we have $f'(x) = x^{-\frac{1}{3}} - 1$ and $f(8) = 4$ find $f(x)$. Ok so $f(x) = \frac{3}{2}x^{\frac{2}{3}} -x + C$. Since $f(8) = 4$ then $f(x) = \frac{3}{2}x^{\frac{2}{3}} -x + 6$

If we have $f''(x) = 2x^{2}, f'(3) = 10, f(3) = 6$ find $f(x)$. So we have $f'(x) = \frac{2}{3}x^{3} - 8$. Now $f(x) = \frac{1}{6}x^{4} - 8x + C$ So would $f(x) = \frac{1}{6}x^{4} - 8x + 16.5$?

Thanks

Last edited: Feb 9, 2005
2. Feb 9, 2005

looks good

3. Feb 9, 2005

### dextercioby

It's correct.

Daniel.

4. Feb 9, 2005

### dextercioby

And the second one is correct as well.

Daniel.

5. Feb 9, 2005

thanks a lot

Just had a few more questions:

An object is projected upwards from the ground with an iniital velocity of 80 feet per second.

(a) How long does it take the object to reach its maximum height?
(b) What is the maximum height?
(c) When is the velocity of the object half of its inital velocity?
(d) What is the height of the object when its velocity is one-half the initial velocity?

Ok so I assume that the acceleration is $-32 \frac{ft}{sec}$ So for (a) $v = -32x + C$ where v is the velocity. For all of these parts do I just take the integral to get back the original function and solve for the required variables?

Also if the rate of change of revenue ie modeled by $\frac{dR}{dt} = 0.675t^{\frac{3}{2}}, 0\leq t\leq 225$ then when $t = 0, R = 0$ if I want to find the model for the revenue function would i simply take the integral of it? Then to see if its at a particular value I just solve for it right?

Thanks

6. Feb 9, 2005

### dextercioby

Pay attention with the first problem:it's v=v(t),not of "x".

Then for the second problem,yes,u'd have to integrate the derivative.

Daniel.

7. Feb 9, 2005

Yea I think I got it

(a) 2.5
(b) 100
(c) 1.25
(d) 75

8. Feb 9, 2005

### xanthym

Your answers look good. However, you should always try to present them with their UNITS. For instance, item "a" should be "2.5 sec".