Question

What to do to solve this problem?
find the difference in quotient of F; [f(x+h)-f(x)]/h; h can't equal 0.

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

Answer 1

Find the derivative by first principle?

Answer 2

all i know is that it equals 8x+4h+5 but i'm not sure how it's solved =[

Answer 3

That's the formula for derivatives. Do you want me to do this for you and we'll see what happens?
Are you sure it's "As h approaches 0", or is it definantly "h can't equal 0"

Answer 4

easy,
so, F(x) = x^2 - x + 4
F(x+h) = (x+h)^2 -(x+h) +4

Now, the formula states, [f(x+h)-f(x)]/h
so,
=(((x+h)^2 -(x+h) +4) - (x^2 - x + 4)))/h
Expand
=((x^2+2xh + h^2 - x + h + 4) - (x^2 - x + 4))/h
x^2 is gone
x is gone
4 is gone
This leave
=(2xh + h^2 + h)/h
factor out h
=h(2x+h+1)/h
h cancels out
=2x+h+1
As h approaches 0
=2x+0+1
=2x+1

Answer 5

it says h≠0. I'm just not sure what the equation give is for or how its used to solve the function. Any ideas?

Answer 6

That's the derivative by first principle

Answer 7

Should be 2x-1 rather than 2x+1. Don't ask me where Spykez went wrong.

Answer 8

i see where i went wrong, it's -x-h, not -x+h

Answer 9

so what is h or where did it come from?

Answer 10

the derivative is a formula for the slope of the line tangent to the curve. your curve is
\[y=x^2-x+4\] suppose you let x = 2 you get y = 6 so the point (2,6) is on the curve. then if you let x = 3 you get 10 so the point (3,10) is on the curve and the slope between the two point is 4

in my example x = 2 and h = 1 because 2+1=3
so i computed for the slope
\[\frac{f(2+1)-f(2)}{1}\] which looks just like
\[\frac{f(x+h)-f(x)}{h}\]

Answer 11

now if you want the slope of the line between (2,6) and (2.1,f(2.1)) you get it by computing
\[\frac{f(2.1)-f(2)}{.1}\] in this case x = 2 and h = .1

Answer 12

by the way people who answered the question were anticipating the derivative, but your question says nothing about letting h go to zero, so it is just pure algebra. your answer would be
\[2x-1+h\]