# secant

• Apr 12th 2010, 03:49 PM
euclid2
secant
Determine an expression, in simplified form, for the slope of the secant PQ

a) $\displaystyle P(1,3), Q(1+h,f(1+h))$, where $\displaystyle f(x)=3x^2$

Now, i know which formula to use, which is

$\displaystyle Limh->0 \frac{(x+h)-f(x)}{h}$

i have the solution to this problem, but don't understand where the numbers are coming from so it's useless to me. i don't understand how to plug the numbers into this formula

for the first step it is:

$\displaystyle Limh->0 \frac{3(1+h)^2-3}{h}$

then simply simplifying to get

3h+6

can someone please explain this to me?
• Apr 12th 2010, 03:56 PM
Quote:

Originally Posted by euclid2
Determine an expression, in simplified form, for the slope of the secant PQ

a) $\displaystyle P(1,3), Q(1+h,f(1+h))$, where $\displaystyle f(x)=3x^2$

Now, i know which formula to use, which is

$\displaystyle Limh->0 \frac{(x+h)-f(x)}{h}$

i have the solution to this problem, but don't understand where the numbers are coming from so it's useless to me. i don't understand how to plug the numbers into this formula

for the first step it is:

$\displaystyle Limh->0 \frac{3(1+h)^2-3}{h}$

then simply simplifying to get

3h+6

can someone please explain this to me?

Formula should be $\displaystyle \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

You're taking $\displaystyle x=1$ and hence $\displaystyle 1+h$.

So $\displaystyle f(1) = 1$ and $\displaystyle f(1+h) = 3(1+h)^2$.

Hence, $\displaystyle \lim_{h \to 0} \frac{f(1+h)-f(1)}{h} = \frac{3(1+h)^2 - 3}{h}= \frac{3h^2 + 6h}{h} = 3h + 6 = 6$ as $\displaystyle h \to 0$
• Apr 12th 2010, 04:15 PM
euclid2
Quote:

Formula should be $\displaystyle \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

You're taking $\displaystyle x=1$ and hence $\displaystyle 1+h$.

So $\displaystyle f(1) = 1$ and $\displaystyle f(1+h) = 3(1+h)^2$.

Hence, $\displaystyle \lim_{h \to 0} \frac{f(1+h)-f(1)}{h} = \frac{3(1+h)^2 - 3}{h}= \frac{3h^2 + 6h}{h} = 3h + 6 = 6$ as $\displaystyle h \to 0$

so what you're doing is plugging (1+h) into 3x^2 to get 3(1+h)^2 ??

and thanks for the help!!
• Apr 12th 2010, 04:22 PM
Quote:

Originally Posted by euclid2
so what you're doing is plugging (1+h) into 3x^2 to get 3(1+h)^2 ??

and thanks for the help!!

Yeah, see what I think is going on in the question is you are being asked to find the slope of the curve at the point (1,3).

Now you can do this simply by differentiating f(x) (= 6x) and plugging in x=1.

But it seems they want you to use the formula...

$\displaystyle \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$