Expert: Alon Mandes - 5/25/2009

Question
Hi so I need help figuring out the answers to problems I did that I got wrong. I just want to know where I went wrong with answers.


Some of the questions include:

1b) 5/x^2 - (^3sqrt(x))        *the square root is x^(1/3)*

I got 2 out of 3 points on this and got the answer of
-5(2x)/(x^2)^2
Could you show me steps and the correct answer if mine is wrong.

~*~*~*~
2. There is a graph that looks like this:
x-ints at -2 and 3. there is a point at (-1,2) that has a sharp angle that goes through both sides to the x-ints (if that helps)

The question for this problem was:
Suppose f is given by the graph below, and g is a differentiable funvtion with the following values:
g(1)= -2
g'(1)=4
g'(-1) =3

a) if h(x)=f(x)g(x), find h'(1)
for this i got (1)(-2)+(1)(4)
                =2
I guess 1 doesn't work here??

b) if k(x)=f(g(x)), find k'(1).
This i completely got wrong.

I couldn't figure out what I was suppose to do.



One more if that is ok..

3a) It says find the derivative of f(x)= x-1
                                       -----
                                       1-x^2
I got the answer of f'(x)=(x^2-2x+1)/((1-x^2)^2)

I only got 7/8 points. What did I do wrong??

Answer
Section 3 is 100% correct ! you shoud review your math teacher.
In Section 1 I didn't understand what you were asked to do, sorry!
Section 2:
(a) is wrong , because the function f is f(x)=(3/2)-x/2,
Thus, if h(x)=f(x)g(x) then h'(x)=f'(x)g(x)+f(x)g'(x), hence,
h'(1)=f'(1)g(1)+f(1)g'(1) = (-1/2)(-2)+[(3/2)-(1/2)]4 = 5 .
(b) If k(x)=f(g(x)), then k'(x)=f'(g(x))*g'(x), hence ,
k'(1)=f'(g(1))*g'(1) = f'(-2)*4. Here f(x) is : f(x)=4+2x,
Thus : k'(1)=(2)*(4)=8.
Note that , the 2 diffrent sharp beams comming from the point
(-1,2), are two diffrent linear function for f, each f for diffrent
domain .

Alon.