Ok, this is probably something easy...

[Lizzie]

Problem:
Find f(x) if:

a. f'(x)=x³+12x+3 given that f(1)=6

b. f"(x)=x+x^1/2 given that f'(1)=0

c. f"(x)=x²+3 cos(x) given that f'(0)=3

What I'm Thinking . . . :

This seems like once I understand a small piece of information then I will easily be able to do it. I just have to figure out what's going on here. I am going to look in my book and online for some help, but you guys usually help me out most, so any help would be very much appreciated!
-Thanks!

 

[galactus]

Hey Lizzie. I believe what you need to do is integrate your derivative. That will

give you f(x). You have a condition that f(1)=6. You must tack on a constant

onto your f(x) so that when x=1, you get 6 as the answer, plus the derivative of

this finction must be what you were given. Clear as mud?.

I'll do 'a' so you can see.

\(\displaystyle f'(x)=x^{3}+12x+3\)

Now integrate:

\(\displaystyle \int((x^{3}+12x+3)dx\)

\(\displaystyle =\frac{x^{4}}{4}+6x^{2}+3x=f(x)\)

Now, you must add something to the end which when x=1, you get 6. Since the

derivative of a constant is 0, when you differentiate, you'll get the

derivative you were given.

\(\displaystyle \frac{x^{4}}{4}+6x^{2}+3x+C=6\)

\(\displaystyle \frac{(1)^{4}}{4}+6(1)^{2}+3(1)+C=6\)

\(\displaystyle \frac{37}{4}+C=6\)

\(\displaystyle C=\frac{-13}{4}\)

So, the equation you need is:

\(\displaystyle \frac{x^{4}}{4}+6x^{2}+3x-\frac{13}{4}\)

Plug in 1 and check.

Now, take the derivative and see if you get what you were given.

I think you'll will.

 

[stapel]

Have you learned the integration rules (similar to the differentiation rules) yet? If so, just apply them to integrate each function.

Remember that indefinite integrals come with a constant of integration, "C". Then evaluate your results at the given x-values, plugging in the given y-values for "f(x)". Solve the resulting linear equation for the value of C.

Eliz.

Edit: Too fast for me, galactus! :wink:

 

[galactus]

Sorry, Stapel. I have a feeling Lizzie will be with us again,

 

[Lizzie]

OK, don't have much time here, but i think I understand! Thanks guys!