For which value of $x$ is the average rate of change equal to the instantaneous rate of change? The average rate of change for $f(x)=x^2+4x-6$ on the interval $[1,3]$ is $8$.
I am not interested in final answer but more how to get there. I am going through calculus right now and already know about derivatives and rate of change. My problem is how to get this word problem into math language and try to solve it.
Inst. rate of change is derivative when lim approaches $0$
average $f(x+h)-f(x)$ divided by $h$.
 A: The average rate of change of $f(x)$ over the interval $a \le x \le b$ is given by
$$\frac{f(b)-f(a)}{b-a}$$
The instantaneous rate of change is given by the derivative $f'(x)$.
In your case, $a=1$ and $b=3$, and so you need to find $1\le x \le 3$ for which
$$f'(x) = \frac{f(3)-f(1)}{3-1}$$
A: The question is asking for what value of $x$ will the average rate of change of our function $f$ be equal to the instantaneous rate of change. In other words, for what value of x will make this expression true:
$$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\frac{f(b)-f(a)}{b-a}$$
We start off by first solving for our average rate of change given the interval $1\le x\le3$:
$$A=\frac{f(b)-f(a)}{b-a}=\frac{f(3)-f(1)}{3-1}=\frac{16}{2} = 8$$
Next we find the derivate of our function $f$ using our limit definition of the derivative:
$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{(x+h)^2+4(x+h)-6-(x^2+4x-6)}{h}$$
$$=\lim_{h\to 0}\frac{x^2+2hx+h^2+4x+4h-6-x^2-4x+6}{h}$$
$$=\lim_{h\to 0}\frac{h^2+2hx+4h}{h}$$
$$=\lim_{h\to 0}\frac{h(h+2x+4)}{h}$$
$$=\lim_{h\to 0}h+2x+4$$
$$=2x+4$$
Finally we set $f'(x)=A$ and solve algebraically:
$$8=2x+4$$
$$4=2x$$
$$x=2$$
We now see that at $x=2$ our instantaneous rate of change is equal to the average rate of change.
