Cal Test tom, few questions

1. Jul 9, 2006

peanutplanters

Im cramming for my cal test tomorrow, and im having a brain fart right now.

[10]
Heres a few problems on my Cal review i need some help with:

If f(5)=3 and f'(5)=2, use differentials to approx. f(5.01) - no clue what to do, i just need a point in the right direction

[7]
Let f(x) = x^3+2x^2-1 Suppose x = 2 and dx = 0.02

A) Calculate delta y
B) Calculate dy - already done

I solved part b using the equation dy=f'(x)dx but im stuck on part a

Last edited: Jul 9, 2006
2. Jul 9, 2006

Gagle The Terrible

no.10 : use the definition of the derivative and use delta x = 0.01

no.7 : a ) Here again, use the definition of the derivative
b ) great !
Hope it helps

3. Jul 9, 2006

d_leet

For the second question I think it's just asking for the difference between the y values at those two points in other words the numerical difference between f(x + dx) and f(x).

For the first question you should know that you can approximate a function around any point where you know the value of the function at that point and the slope of the tangent line at that point, what you're doing is using the tangent line to the function as an approximation to the actual function.

The derivative is defined as the limit of (f(x + h) - f(x))/h as h goes to zero. Now for small h but nonzero h we should be able to say that

f'(x) is approximately equal to (f(x+h) - f(x))/h

and since you have f(5) and f'(5) you can approximate f(5.01).

4. Jul 9, 2006

peanutplanters

i got question #7

for question #10 however,

f(5)=3 f'(5)=2 approx. f(5.01)=?

im still confused how do my two equations help me solve the problem?

delta y = f(x+delta x) - f(x)
dy = f'(x)dx

Last edited: Jul 9, 2006
5. Jul 9, 2006

d_leet

Ok using those equations you have that f(x + delta x) - f(x) = delta y

and dy = f'(x)dx

But for relatively small x delta y and dy are approximately equal so we can substitute the second equation into the first giving

f'(x)dx = f(x + delta x ) - f(x)

so you approximate f(x + delta x ) by the equations above as
f(x + delta x ) = f(x) + f'(x)*(delta x) because we are assuing that for small x dx is approximately equal to delta x.

6. Jul 9, 2006

peanutplanters

Thanks for all the help, almost finished:

14. Find the function f that satisfies the following conditions:

f ''(x) = sin x , f'(0)=2 , f(pi) = -1

I understand u can take the anti deravitive of f ''(x) and then solve for C then, take the anti deravitive of that ( f(x) ). But im stuck because there isnt an anti deravitive of sin x. Am i setting this problem up wrong?

7. Jul 9, 2006

d_leet

What is the derivitave of cos(x)? How about the derivative of -cos(x)?

8. Jul 9, 2006

peanutplanters

so -f'(x) = cos x ? therefore f'(x) = -cos x?

9. Jul 9, 2006

d_leet

Well since you started with f"(x) = sin(x) you technically have f'(x) = -cos(x) +C, and since you know f'(0) =2 you can solve for C then integrate again and repeat the same process to find f(x).

10. Jul 9, 2006

nrqed

Why do you say there is no antiderivative of sinx? It's simply -cos(x) +C !
Then you can integrate this again.

11. Jul 9, 2006

peanutplanters

ahh that makes sense, thanks alot. both of you thanks

12. Jul 9, 2006

d_leet

Glad to help, and it's good to hear that it makes sense now. Good luck on your test tomorrow.