# Math Help - Average rate of change

1. ## Average rate of change

I am unsure as to which forum to put this in. I took a shot and put it here, apologies if wrong.

Okay, to the question I am having trouble with.

"Find the average rate of change of $f(x) = \sqrt{x + 11}$ with respect to x from $x = 5$ to $x = 5 +h$. [4 marks]"

$f(5) = \sqrt{5 + 11}$
$f(5) = \sqrt{16}$
$f(5) = 4$

$f(5 + h) = \sqrt{5 + h + 11}$
$f(5 + h) = \sqrt{16 + h}$
$f(5 + h) = 4 + \sqrt{h}$
*I am unsure if I am correct in square rooting the 16, so having 4 + the square root of h.*

Then I use the $\frac{y2 - y1}{x2 - x1}$ formula to get the slope of the secant *rate of change*:

$\frac{4 + \sqrt{h} - 4}{5 - 5 + h}$
$= \frac{\sqrt{h}}{h}$
That's my final solution. My problem with this solution is when I sub a number in for h, I get a different answer from running the numbers through, and simply going *square root of the number divided by the number*.
Example:
Let's pretend h is 20.
So I have $f(5) = 4$.
I can sub in 20 for h:
$f(5 + 20) = \sqrt{5 + 20 + 11}$
$f(25) = \sqrt{36}$
$f(25) = 6$

Run that through the slope formula:
$\frac{6 - 4}{20 - 5}$
$= \frac{2}{5}$
$= 0.4$

Then, using my average rate of change thing: $\frac{\sqrt{h}}{h}$ I get:
$\frac{\sqrt{20}}{20}$
$= 0.223606797$
which is a completely different answer. So I must be doing something wrong here.

Okay, I tried using 25 for h in the $\frac{\sqrt{h}}{h}$ and it gives the same ansewr as running the math through. So it is not technically h, but it is h + 5, so is this actually right??? I see that as: $\frac{\sqrt{h+5}}{h+5}$ which I think is different, please correct me, I am so confused.

EDIT:There, I have successfully turned this into something readable using latex

2. Originally Posted by Kakariki
I am unsure as to which forum to put this in. I took a shot and put it here, apologies if wrong.

Okay, to the question I am having trouble with.

"Find the average rate of change of $f(x) = \sqrt{x + 11}$ with respect to x from $x = 5$ to $x = 5 +h$. [4 marks]"

$f(5) = \sqrt{5 + 11}$
$f(5) = \sqrt{16}$
$f(5) = 4$

$f(5 + h) = \sqrt{5 + h + 11}$
$f(5 + h) = \sqrt{16 + h}$
$f(5 + h) = 4 + \sqrt{h}$
*I am unsure if I am correct in square rooting the 16, so having 4 + the square root of h.*

Then I use the $\frac{y2 - y1}{x2 - x1}$ formula to get the slope of the secant *rate of change*:

$\frac{4 + \sqrt{h} - 4}{5 - 5 + h}$
$= \frac{\sqrt{h}}{h}$
That's my final solution. My problem with this solution is when I sub a number in for h, I get a different answer from running the numbers through, and simply going *square root of the number divided by the number*.
Example:
Let's pretend h is 20.
So I have $f(5) = 4$.
I can sub in 20 for h:
$f(5 + 20) = \sqrt{5 + 20 + 11}$
$f(25) = \sqrt{36}$
$f(25) = 6$

Run that through the slope formula:
$\frac{6 - 4}{20 - 5}$
$= \frac{2}{5}$
$= 0.4$

Then, using my average rate of change thing: $\frac{\sqrt[h]][h]]$ I get:
$\frac{\sqrt{20}}{20}$
$= 0.223606797$
which is a completely different answer. So I must be doing something wrong here.

Okay, I tried using 25 for h in the $\frac{\sqrt{h}}{h}$ and it gives the same ansewr as running the math through. So it is not technically h, but it is h + 5, so is this actually right???

the wrong thing as what you said you can't separate 16 from the square root

$f(5+x)=\sqrt{16+h}$ you can't simplify it more than that

$s=\frac{\sqrt{16+h} - 4 }{h}$\

EDIT:There, I have successfully turned this into something readable using latex
CONGRATULATIONS LATEX is simple and useful