<s>[INST] What is the Rate of change of f (x^2) given rate of change of f (x) Find the average rate of change of $f(x^2)$ on the interval [1,4] given that  the average rate of change of $f (x)$ equals 9
on  interval [1,16]?
This question  has two different "answers" according  to  two different  teachers 
The first  give answer of 9 by assuming $y=x^2$ and applying the formula  of rate of change as following 
Let$ y=x^2 $ with $I$= [1,4] then$ f (1)$=1,$f (4)$=16 
$\frac{f(4^2) - f(1^2)}{(4^2 -1^2)}$  = 
$\frac{f (y_{2})-f (y_{1})}{y_{2}-y_{1}}$
 =
$\frac{f(16) - f(1)}{(16-1)}$ =9
The second give the answer of 45 
as following 
$\frac{f(b) - f(a)}{(b - a)}$ to $\frac{f(16) - f(1)}{16 - 1}=9$ to 
$\frac{f(16) - f(1)}{15}=9$   
$f(16) - f(1)=9*15$
Now  we find rate of change of $f (x^2)$ following 
$\frac{f(16) - f(1)}{4 - 1}=\frac{9*15}{3}=45$
what is
the  right  answer?
Can we ensure  the either answers geometrically? 
Thank you for helping 
 [/INST] Let $g(x) = f(x^2)$. 
The average rate of change of a function $f$ on an interval $[a,b]$ is $\dfrac{f(b) - f(a)}{b-a}$.  
The average rate of change of $f$ on the interval $[1,16]$ is $$\frac{f(16) - f(1)}{16 - 1} = \frac{f(16) - f(1)}{15}.$$
You are told this rate of change equals $9$.
The average rate of change of $g$ on the interval $[1,4]$ is thus equal to $$ \frac{g(4) - g(1)}{4-1} = \frac{f(16) - f(1)}{3} = 5 \cdot \frac{f(16) - f(1)}{15} = 45.$$
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