Suppose that $\displaystyle g(0) = -7$ and that $\displaystyle g'(t)=5$ for all of t. Must $\displaystyle g (t) = 5t-7$ for all of t??
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 $\displaystyle y= 2x-7$, then $\displaystyle y^{-1} = \frac {x+7}{2}$. when x=0, y=-7, but when $\displaystyle y^{-1} = 5, x = -2$ so that negates the question. Is my thought process skewed? Or is that a valid approach?
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.