# Let $f (x) = 2^x$. Show that…

Let $f(x) = 2^x$. Show that $\dfrac{f(x+h) - f(x)}{h} = \dfrac{2^x(2^h-1)}{h}$.

First day of my precalc class in college, and I have no idea where to start to solve this one. Can anyone point me in the right direction?

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Can you find $f(x+h)$? Are you sure you mean $\dfrac{f(x+h) + f(x)}{h}$ rather than $\dfrac{f(x+h) - f(x)}{h}$? –  JavaMan Sep 11 '12 at 22:08
Don't you mean $f(x+h)-f(x)$? –  Per Manne Sep 11 '12 at 22:11
Do you know how to evaluate a function at a point at all? If $f(x) = \text{ expression in } x$. Then $f(x+h) = \text{ the same expression, but in } x+h$. Also, $2^{x\pm h} = 2^{x} \cdot 2^{\pm h}$. –  user2468 Sep 11 '12 at 22:14
I'm guessing that plus outside the parentheses should be a minus. Also, you'll be seeing that slope equation on the left side again once you get to calculus. –  Mike Sep 11 '12 at 22:15
Nominated for Today's Most Incipit Question Title. –  Henning Makholm Sep 11 '12 at 23:46

Start by plugging in what you know. You're trying to figure out what $(f(x+h) - f(x))/h$ is equal to, and you know what $f(x)$ equals. So why don't you plug in all that and see what you get. From there, just remember the exponent formula $x^{a +b} = (x^a)(x^b)$.

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thank you,this problem is very simple when reading neat text. On my binder paper I had written f(x)+h instead of the correct f(x+h). –  Thumbtack Sep 11 '12 at 23:37

Let $f(x)=2^x$, then $f(x+h)$ is the same that replace the original function by $x+h$, then $f(x+h)=2^{(x+h)}$, which by basic school algebra is $2^x$ times $2^h$, then, you can replace your function $\dfrac{f(x+h) + f(x)}{h}$ by $\dfrac{(2^x)(2^h)+2^x}{h}$ if you factorize it, you get $\dfrac{2^x(2^h+1)}{h}$ which is what you nedded, then you get your answer.

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$\frac{f(x+h)-f(x)}{h}=\frac{2^{(x+h)}-2^x}{h}$using laws of exponentiation we know that for real numbers a,b and c $a^b*a^c=a^{(b+c)}$using this we get that $\frac{2^{(x+h)}-2^x}{h}=\frac{2^x(2^h-1)}{h}$

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I don't understand the downvote for this one. –  Ross Millikan Sep 11 '12 at 23:34

You have $f(x) = 2^x$. So for example you would have $f(2) = 2^2$ and $f(7) = 2^7$. So, when you evaluate $f$ at some number, then you just put that number in where the $x$ is.

You also have for example $f(x + h) = 2^{x+ h}$. So:

$$\dfrac{f(x+h) - f(x)}{h} = \dfrac{2^{x + h} - 2^x}{h}.$$

All you do then is factor out an $2^x$ in the numerator (which I will assume that you know how to do).

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I don't understand the downvote for this one. –  Ross Millikan Sep 11 '12 at 23:34