Question

Solve for the function x^2 + 2x - 4, find lim h approaching 0 = f(h+4) - f(4)/h

Answer 1

I have attached an image of the equation if that is of any assist.

Answer 2

\[f(h+4)=(h+4)^2+2(h+4)-4\]
\[f(4)=4^2+2(4)-4\]
So we have
\[\lim_{h \rightarrow 0}\frac{(h+4)^2+2(h+4)-4-(4^2+2(4)-4)}{h}\]

Answer 3

\[\frac{f(4+h) - f(4)}{h} = \frac{((4+h)^2 + 2(4+h) -4) - (4^2 + 2(4) -4)}{h}\]

now collect like terms

Answer 4

That is how I figured it out prior to posting on open study. I received the answer h^2 + 10h + 2h / h

Answer 5

is that the answer?

Answer 6

since I can't factor the h^2?

Answer 7

\[\lim_{h \rightarrow 0}\frac{h^2+8h+16+2h+8-4-16-8+4}{h}=\lim_{h \rightarrow 0}\frac{h^2+8h+2h}{h}\]

Answer 8

\[\lim_{h \rightarrow 0}\frac{h(h+8+2)}{h}=\lim_{h \rightarrow 0}(h+10)\]

Answer 9

I understand that. But, my teacher taught it as if I had to take f(4) and substitute the 4 throughout the function x^2 + 2x - 4.

This is what I did:

f(4) = 4^2 + 2(4) - 4 = 20

Then, we were to take that number and add it to the h(h + 8 + 2) + 20/h, like so.

Is this incorrect?

Answer 10

Never mind. Thank you for your assistance. You both were a great help. =]