Question

f(x)=x^2+3x-5

Solve
(f(x+h)-f(x))/h

h cannot = 0

Answer 1

bunch of algebra
ready?

Answer 2

let's do this

Answer 3

ok first we need
\[f(x+h)\] and since
\[f(x)=x^2+3x+5\] then
\[f(x+h)=(x+h)^2+3(x+h)+5\] now comes the first part of algebra

Answer 4

square carefully and get
\[f(x)=x^2+2xh+h^2+3x+3x+5\]

Answer 5

now we need
\[f(x+h)-f(x)=x^2+2xh+h^2+3x+3h+5-(x^2+3x+5)\]

Answer 6

sure looks like a derivative, df/dx = 2x + 3, etc.

Answer 7

don't forget the parentheses about the last part
distribute the minus sign and get
\[x^2+2xh+h^2+3x+3h+5-x^2-3x-5\] combine like terms now

Answer 8

@douglaswinslowcooper in particular, it's a difference quotient; I assume that if they're just working with these, they may not know the concept of a limit yet, and hence you can't do derivatives.

Answer 9

everything without an \(h\) will go and you get
\[f(x+h)-f(x)=2xh+3h+h^2\]

Answer 10

finally divide by \(h\) to ge t
\[\frac{f(x+h)-f(x)}{h}=\frac{2xh+3h+h^2}{h}\] divide each term by \(h\) leaving
\[2x+3+h\]

Answer 11

like a said, bunch of algebra, and there is lots of room for mistake here
especially squaring \((x+h)^2\) and also making sure to put \(-f(x)\) in parentheses