Functions

[WTF?]

Hello, I'm in need of another push again...

It says:

Let $\displaystyle g(x)=x^2$

a.$\displaystyle g(x+h)$
b.$\displaystyle g(x+h) - g(x)$
c.$\displaystyle (g(x+h) - g(x))/(h)$

Okay, I'm stuck on all...I'm not entirely sure how to approach this, but would I need to substitute $\displaystyle x^2$? It seems kind of obvious and delicate....

Thanks again for any replies.

 

[soroban]

Hello, WTF?!

"Substitute $\displaystyle x^2$" ? . . . You're forgetting the basics . . .

Let $\displaystyle g(x)\,=\,x^2$

a) Find: $\displaystyle g(x+h)$

Okay, some baby-talk . . .

What does $\displaystyle g(4)$ mean?
$\displaystyle \;\;$It means "Replace $\displaystyle x$ with $\displaystyle 4$".

So we have: $\displaystyle \,g(4)\,=\,4^2\,=\,16$

So what does $\displaystyle g(x+h)$ mean?
$\displaystyle \;\;$It means "Replace $\displaystyle x$ with $\displaystyle x+h$".

So we have: \(\displaystyle \.g(x+h)\:=\x\,+\,h)^2\:=\:x^2\,+\,2xh\,+\,h^2\)

b) Find: $\displaystyle g(x+h)\,-\,g(x)$

What does it say?
$\displaystyle \;\;$It says: Find $\displaystyle g(x+h)$, then subtract $\displaystyle g(x)$.

Well, we just found $\displaystyle g(x+h)$ and we know what $\displaystyle g(x)$ is.
$\displaystyle \;\;$So we have: $\displaystyle \,g(x+h)\,-\,g(x)\;=\;(x^2\,+\,2xh\,+\,h^2)\,-\, x^2\;=\;2xh \,+\,h^2$

c) Find: \(\displaystyle \L\frac{g(x+h)\,-\,g(x)}{h}\)

What does it say?

It says: Take what we found in part (b), and divide by $\displaystyle h$.

So we have: \(\displaystyle \L\,\frac{2xh\,+\,h^2}{h}\)\(\displaystyle \;=\;\L\frac{\not{h}(2x\,+\,h)}{\not{h}}\)$\displaystyle \;=\;2x\,+\,h$