# Math Help - Derivates Question

1. ## Derivates Question

#1. The functions f and g are differentiable. For all x, f(g(x)) = x and g(f(x)) = x. If f(3) = 8 and f'(3) = 9, what are the values of g(8) and g'(8)?

#2. The following values are given:
f(2.5) = 31.25
f(2.8) = 39.20
f(3.0) = 45
f(3.1) = 48.05
For the function f which is differentiable. Both f and f' are strictly increasing on the interval 0 < x < 5. Which of the following could be the value of f'(3)?
(A) 20
(B) 27.5
(C) 29
(D) 30
(E) 30.5
- originally, I tried to approximate f' by using the secant line and taking f(3.1) - f(2.8)/3.1 - 2.8 but ended up getting 29.5 as an answer..

2. Originally Posted by xxlvh
#1. The functions f and g are differentiable. For all x, f(g(x)) = x and g(f(x)) = x. If f(3) = 8 and f'(3) = 9, what are the values of g(8) and g'(8)?
g(f(3))=g(8)

but

g(f(3))=3

so:

g(8)=3.

From the chain rule we have:

g'(f(x))f'(x)=1

So:

g'(f(3))f'(3)=1

CB

3. Originally Posted by xxlvh
#1. The functions f and g are differentiable. For all x, f(g(x)) = x and g(f(x)) = x. If f(3) = 8 and f'(3) = 9, what are the values of g(8) and g'(8)?

#2. The following values are given:
f(2.5) = 31.25
f(2.8) = 39.20
f(3.0) = 45
f(3.1) = 48.05
For the function f which is differentiable. Both f and f' are strictly increasing on the interval 0 < x < 5. Which of the following could be the value of f'(3)?
(A) 20
(B) 27.5
(C) 29
(D) 30
(E) 30.5
- originally, I tried to approximate f' by using the secant line and taking f(3.1) - f(2.8)/3.1 - 2.8 but ended up getting 29.5 as an answer..
As $f$ and $f'$ are strictly increasing we have for all $h,g>0$:

$\frac{f(x)-f(x-h)}{h}>f'(x)>\frac{f(x+g)-f(x)}{g}$

Now use $h=0.2\ x=3\ g=0.1$

CB

4. I'm still a little bit lost...
Well, when I solved for g'(f(3))f'(3)=1 I had g'(8) equalling 1/9. (Hopefully I did that correctly) I'm just not sure where the "1" came from for that equation.
And for question 2..where did the values of h, x, and g come from?

5. Originally Posted by xxlvh
I'm still a little bit lost...
Well, when I solved for g'(f(3))f'(3)=1 I had g'(8) equalling 1/9. (Hopefully I did that correctly) I'm just not sure where the "1" came from for that equation.
we have:

$g(f(x))=x$

diffenetiate both sides to get:

$g'(f(x))f'(x)=\frac{d}{dx}x=1$

CB

6. Originally Posted by xxlvh
And for question 2..where did the values of h, x, and g come from?
The inequality is true for arbitary positive $h$ and $g$, and arbitary $x$.

Now we want to know about the derivative at $x=3$, which determines the value of $x$ of interest. In the tabulated data the previous $x$ is $2.8$ or $0.2$ less than $3$, and the next point is for $3.1$, or $0.1$ greater than $3$.

So we have:

$
\frac{f(3)-f(2.8)}{0.2}>f'(3)>\frac{f(3.1)-f(3)}{0.1}
$

CB