[SOLVED] Inverse functions

• Dec 10th 2009, 06:41 PM
Ife
[SOLVED] Derivatives of functions
Suppose that $g(0) = -7$ and that $g'(t)=5$ for all of t. Must $g (t) = 5t-7$ for all of t??
• Dec 10th 2009, 06:45 PM
VonNemo19
Quote:

Originally Posted by Ife
Suppose that $g(0) = -7$ and that $g'(t)=5$ for all of t. Must $g (t) = 5t-7$ for all of t??

Absolutely not. There are an infinte number of functions that pass through the point $(0,-7)$ having slope 5.
• Dec 10th 2009, 07:18 PM
Ife
Quote:

Originally Posted by VonNemo19
Absolutely not. There are an infinte number of functions that pass through the point $(0,-7)$ having slope 5.

Great. I had the right answer, but i am not sure that i had the right thought process... i had said something along the lines of: If $y= 2x-7$, then $y^{-1} = \frac {x+7}{2}$. when x=0, y=-7, but when $y^{-1} = 5, x = -2$ so that negates the question. Is my thought process skewed? Or is that a valid approach?
• Dec 11th 2009, 02:05 AM
mr fantastic
Quote:

Originally Posted by Ife
Suppose that $g(0) = -7$ and that $g'(t)=5$ for all of t. Must $g (t) = 5t-7$ for all of t??

Yes:

$\frac{dg}{dt} = 5 \Rightarrow g(t) = 5t + C$. $g(0) = 7 \Rightarrow 7 = C$. Therefore $g(t) = 5t + 7$.

Quote:

Originally Posted by VonNemo19
Absolutely not. There are an infinte number of functions that pass through the point $(0,-7)$ having slope 5.

Correct. But not having a slope 5 for all values of t ...
• Dec 13th 2009, 07:10 AM
Ife
Quote:

Originally Posted by mr fantastic
Yes:

$\frac{dg}{dt} = 5 \Rightarrow g(t) = 5t + C$. $g(0) = 7 \Rightarrow 7 = C$. Therefore $g(t) = 5t + 7$.

Correct. But not having a slope 5 for all values of t ...

Thanks, I was reviewing the question yesterday and realised that entire thing before was Wrong! Because we are not looking at the inverse, but rather, the derivative! I was about to make a note of that here. Didn't see you had responded.

Thanks!
• Dec 13th 2009, 07:11 AM
Ife
Quote:

Originally Posted by VonNemo19
Absolutely not. There are an infinte number of functions that pass through the point $(0,-7)$ having slope 5.

This is if we are looking at the inverse function but we are not. The question asked about the derivative! We had the totally wrong idea here... I hope this post doesn't mislead anyone...
• Dec 13th 2009, 07:38 AM
HallsofIvy
Quote:

Originally Posted by Ife
Suppose that $g(0) = -7$ and that $g'(t)=5$ for all of t. Must $g (t) = 5t-7$ for all of t??

Yes. By the mean value theorem, two functions that have the same derivative must differ by a constant:
Let h(x)= f(x)- g(x). f'(t)= g'(t), then h'(x)= 0 for all t. By the mean value theorem, h(b)- h(a)/(b- a)= 0 so h(b)- h(a)= 0 and h(b)- h(a)= 0. That is, h is a constant and so f and g differ by a constant.

Since f(t)= 5t has f'(x)= 5 for all t,any function with derivative always equal to 5 must be of the form g(x)= 5x+ C. In order that g(0)= -7, C must be -7.