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?
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?
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 $x^{a +b} = (x^a)(x^b)$.
$\frac{f(x+h)-f(x)}{h}=\frac{2^{(x+h)}-2^x}{h}$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} $
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.
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).