Finding the instantaneous rate of change of the function $f(x)=-x^2+4x$ at $x=5$ Finding the instantaneous rate of change of the function $f(x)=-x^2+4x$ at $x=5$,
I know the formula for instantaneous rate of change is $\frac{f(a+h)-f(a)}{h}$
I think it's the negative in front of the x that is throwing me the most.
$$\frac{f(h+5)-f(5)}{h}$$
$$\frac{-(h+5)^2+4(h+5)-(-5)}{h}$$
$$\frac{-(h^2+10h+25)+(4h+20)+5)}{h}$$
$$\frac{(-h^2-10h-25)+4h+20+5}{h}$$
$$\frac{-h^2-6h}{h}$$
$$\frac{h(-h-6)}{h}$$
$$-h-6$$
$0-6=-6$; The instantaneous rate of change is $-6$
Have I done this correctly? I've just started with this stuff and want to make sure I'm not making errors before continuing.
 A: The rate of change of $f$ in the point $x=5$ will be the derivative of $f$ in $x=5$.
You have two ways of doing that (that are the same in essence, you can show it):
$$f'(a) = \lim_{h\rightarrow0} \frac{f(a+h)-f(a)}{h}$$
is equivalent to
$$f'(a) = \lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}$$
Using the first method:
$$f'(5) = \lim_{h\rightarrow0} \frac{f(5+h)-f(5)}{h}$$
$$=\lim_{h\rightarrow0} \frac{-(5+h)^2+4(5+h)-25+20}{h} = $$
$$\lim_{h\rightarrow0} \frac{-25 - 10h -h^2+20+4h+25-20}{h} = \lim_{h\rightarrow0} \frac{-h^2-6h}{h} = $$
$$\lim_{h\rightarrow0} -h - 6 = -6$$
So, yes, your answer is correct. Don't forget to use the limit.
You could use the same definition to find the rate of change for every point $a$.
$$f'(a) = \lim_{h\rightarrow0} \frac{f(a+h)-f(a)}{h} = \lim_{h\rightarrow0} \frac{-(a+h)^2+4(a+h)+a^2-4a}{h} = $$
$$\lim_{h\rightarrow0} \frac{-a^2-2ah-h^2+4a+4h+a^2-4a}{h} = \lim_{h\rightarrow0} \frac{-h^2-2ah+4h}{h} = $$
$$\lim_{h\rightarrow0} -h - 2a +4= 4 - 2a$$
So, in the point $x=5$ the rate of change is $4-2*5 = 4-10 = -6$.
A: Your final answer is right, so well done.
The only minor detail is the notation. The instantaneous rate of change, i.e. the derivative, is expressed using a limit.
$$\mathrm{f}'(x) = \lim_{h \to 0} \left(\frac{\mathrm{f}(x+h)-\mathrm{f}(x)}{h}\right)$$
You need the limit notation on the left of all of your expressions, i.e.
$$\begin{eqnarray*}
\mathrm{f}'(5) &=& \lim_{h \to 0}\left(\frac{\mathrm{f}(5+h)-\mathrm{f}(5)}{h}\right) \\ \\
&=& \lim_{h \to 0} \left( \frac{\left[-(x+h)^2+4(x+h)\right] - \left[-x^2+4x\right]}{h} \right) \\ \\
&\vdots&  \\ \\
&=& \lim_{h \to 0} \left( -h-6\right) \\ \\
&=& -6
\end{eqnarray*}$$
A: The instantaneous rate of change of a function $f(x)$ at $x=a$ is simply given by its derivative at $x=a$, i.e., $f'(a)$. Since $f'(x) = -2 x + 4$, then $f'(5)$ is the answer you are looking for which, by the way, matches your result.
Precisely, you should have written (note this is a limit):
$$IRC_{f(x),x=5} = \lim_{h \to 0} \left.\frac{f(x+h)-f(x)}{h}\right|_{x=5} = f'(5) = \ldots = -6. $$
Cheers!
A: Yes, that is the correct answer. You should add the limit notation to your function as this is merely the limit definition of the derivative. I am assuming you are in a Calc 1 class and thus have not learned the power rule or anything.
