# Thread: Finding and evaluating difference quotient

1. ## Finding and evaluating difference quotient

Morning Forum any help with the following would be appreciated im having issues with it
Find the difference quotient for y = f(x) = 3x2 + 2, and evaluate it for
(a) h = 2 (b) h = 0.2 (c) h = 0.02)
(d) Describe what will happen as h gets smaller and smaller

Thanks - AC-

2. $\displaystyle f(x) = 3x^2 + 2$

$\displaystyle f(x+h) = 3(x+h)^2 + 2$

difference quotient ...

$\displaystyle \frac{f(x+h) - f(x)}{(x+h) - h}$

I would start by expanding the expression for f(x+h)

3. Originally Posted by AlgebraicallyChallenged
Morning Forum any help with the following would be appreciated im having issues with it
Find the difference quotient for y = f(x) = 3x2 + 2, and evaluate it for
(a) h = 2 (b) h = 0.2 (c) h = 0.02)
(d) Describe what will happen as h gets smaller and smaller

Thanks - AC-
Do you know what the difference quotient is? It's a means of approximating the gradient of a point on a curve.

I'll leave you to do the research as to why, but the difference quotient is given by

$\displaystyle \frac{f(x + h) - f(x)}{h}$.

$\displaystyle f(x + h) = 3(x + h)^2 + 2 = 3(x^2 + 2xh + h^2) + 2 = 3x^2 + 6xh + 3h^2 + 2$

$\displaystyle f(x) = 3x^2 + 2$.

$\displaystyle f(x + h) - f(x) = 6xh + 3h^2$

$\displaystyle \frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2}{h} = 6x + 3h$.

Now evaluate this at your given values of h and tell me what happens as h gets closer and closer to 0.

4. Originally Posted by skeeter
$\displaystyle f(x) = 3x^2 + 2$

$\displaystyle f(x+h) = 3(x+h)^2 + 2$

difference quotient ...

$\displaystyle \frac{f(x+h) - f(x)}{(x+h) - h}$

I would start by expanding the expression for f(x+h)
It's actually $\displaystyle \frac{f(x + h) - f(x)}{(x + h) - x}$.