# Thread: find values of the functions

1. ## find values of the functions

just wanted to check if im correct and also wanted to know how i would do the last two...d and e

I have the answers for the first 3...
a)(f-g)'(0) $\displaystyle f'(0) - g'(0) = 1 - 3 = -2$
b)(fg)'(1) $\displaystyle f'g + fg' = (2*4)+(3*1) = 11$
c) (f/g)'(2)$\displaystyle (f'g-g'f)/g^2 = ((3*4)-(4*4))/(2^2) = -1$

just wanted to know if im correct for those ones...and i dont know how to do the conposite functions of of d and e..how would i go on about doing those??

2. hint:

d) $\displaystyle (g \circ f)'(x) = D_x[g(f(x))] = g'(f(x)) \cdot f'(x)$

but i dont know what $\displaystyle g^2$ in e means.. or did it mean square of g? Ü

3. thanks for that hint on question d. but does anyone else know how to do question e?? which is $\displaystyle (f o (g^2))'(4)$

4. Originally Posted by b00yeah05
thanks for that hint on question d. but does anyone else know how to do question e?? which is $\displaystyle (f o (g^2))'(4)$
The clue is in kalagota's hint for (d). Did you carefully follow his/her reasoning and understand it ......?

$\displaystyle (f \circ g^2)'(x) = D_x[f(g^2(x))] = f'(g^2(x)) \cdot (g^2(x))' = f'(g^2(x)) \cdot 2 g(x) \cdot g'(x)$.

Then substitute x = 4.

5. yes i understand it. its the chain rule..now just wondering,,,where you have put the dot sign..do i need to do more work and make it the product rule or is that simply a multiplication sign?? so shall i just substitute the values ??

6. Originally Posted by b00yeah05
yes i understand it. its the chain rule..now just wondering,,,where you have put the dot sign..do i need to do more work and make it the product rule or is that simply a multiplication sign?? so shall i just substitute the values ??
The dots represent multiplication.

7. just wanting to check my answers....

I have the answers for the first 3...
a)(f-g)'(0) $\displaystyle =f'(0) - g'(0) = 1 - 3 = -2$
b)(fg)'(1) $\displaystyle =f'g + fg' = (2*4)+(3*1) = 11$
c) (f/g)'(2) = $\displaystyle (f'g-g'f)/g^2 = ((3*2)-(4*4))/(2^2) = -2.5$
d) $\displaystyle (g o f)'(3) = 0$
e) $\displaystyle (f o (g^2))'(4) = 192$

am i correct?? the values i got are from the table

8. Originally Posted by b00yeah05
just wanting to check my answers....

I have the answers for the first 3...
a)(f-g)'(0) $\displaystyle =f'(0) - g'(0) = 1 - 3 = -2$ Mr F says:

b)(fg)'(1) $\displaystyle =f'g + fg' = (2*4)+(3*1) = 11$ Mr F says:

c) (f/g)'(2) = $\displaystyle (f'g-g'f)/g^2 = ((3*2)-(4*4))/(2^2) = -2.5$Mr F says:

d) $\displaystyle (g o f)'(3) = 0$ Mr F says:

e) $\displaystyle (f o (g^2))'(4) = 192$ Mr F says:

am i correct?? the values i got are from the table

d) $\displaystyle (g \circ f)'(x) = g'(f(x)) \cdot f'(x)$.

So $\displaystyle (g \circ f)'(3) = g'(f(3)) \cdot f'(3)$.

From the table, f(3) = 2.

So $\displaystyle (g \circ f)'(3) = g'(2) \cdot 3 = 4 \cdot 3 = 12$ .....?

------------------------------------------------------------------------

e) $\displaystyle (f \circ g^2)'(x) = f'(g^2(x)) \cdot 2 g(x) \cdot g'(x)$.

So $\displaystyle (f \circ g^2)'(4) = f'(g^2(4)) \cdot 2 g(4) \cdot g'(4)$.

From your table: g(4) = 2 therefore $\displaystyle g^2 (4) = 2^2 = 4$.

So $\displaystyle (f \circ g^2)'(4) = f'(4) \cdot 2(2) \cdot 4 = 3 \cdot 4 \cdot 4 = 48$ .....?

9. so what you have done is got the composite functions and then substituted one value to get another value and used that value from the table and got the answer?? because i thought it was just multitplictaion which i substitute individual values for each..

not trying to question you here but....are you sure what you have done is right because we are told the x value so wouldnt we just hold that a constant?? and use those values from the table??

10. Originally Posted by b00yeah05
so what you have done is got the composite functions and then substituted one value to get another value and used that value from the table and got the answer?? Mr F says: Yes.

because i thought it was just multitplictaion which i substitute individual values for each.. Mr F says: No no.

not trying to question you here but....are you sure what you have done is right because we are told the x value so wouldnt we just hold that a constant?? and use those values from the table?? Mr F says: Questioning is fine. That's how you learn. The answer to your question is no. I'll show you why in my main reply
Consider $\displaystyle h(x) = (x - 1)^2$. Obviously h(3) = 4.

You can think of h(x) as f(g(x)) where g(x) = x - 1 and f(x) = x^2.

Then h(3) = f(g(3)) ......

Now, g(3) = 2. So h(3) = f(g(3)) = f(2) = 2^2 = 4. See how it works?

If you're still not sure, show me all your working for (d) and I'll try and explain where you've gone wrong in your working. Hopefully (e), by far the trickier one, will then make sense.

11. i get the idea of what you have done...but i dont understand why....i know that you used....for eg...the value for g(f(x)) at x = 4 is 1 so it would be
g(f(4)) = g(1) <--- then it changes and now u need to find the value for g(1)???

sorry if that didnt make sense but i dont think i can explain it any better then that. im running out of words. my workin for d previously was...

g(f(x)) = g'(f(x)) . f'(x)
= 0(2) . (3)
= 0

12. Originally Posted by b00yeah05
i get the idea of what you have done...but i dont understand why....i know that you used....for eg...the value for g(f(x)) at x = 4 is 1 so it would be
g(f(4)) = g(1) <--- then it changes and now u need to find the value for g(1)???

sorry if that didnt make sense but i dont think i can explain it any better then that. im running out of words. my workin for d previously was...

g(f(x)) = g'(f(x)) . f'(x)
= 0(2) . (3)
= 0
You need to go back and thoroughly review the concept of a composite function. g'(f(3)) does NOT mean g'(3) times f(3)!!

It means get the value of f(3) and find g' for that value. f(3) = 1, so you want the value of g' at 1. That is, you want g'(1).

You must go back and review composite functions .......

Did you follow my previous example: The h(x) = (x - 1)^2, g(x) = x - 1, f(x) = x^2, h(3) = f(g(3)) business .......

f(g(3)) does NOT mean f(3) g(3) ...... If you did this, you'd get h(3) = 18 instead of 4 ......

Review composite functions. That concept is at the very heart of all your troubles.