Slope of a tangent line and weird world problem

**superfly8912** · August 23rd 2009, 07:20 PM

Three problems:

1) find the slope of the tangent line for the graph f(x)= x^2 at (2,4).

2) Find the slope of the line joining (2,4) and (2+h, f(2+h)) in terms of the nonzero number h. (This is also for the graph f(x)= x^2)

And

A rancher has 300 feet of fence to enclose two adjacent pastures. (See figure) Express the total area A of the two pastures as a function of x. What is the domain of A?

Graph the area function (I suppose I can do this once I find the equation) and estimate the dimensions that yield the maximum amount of area for the pastures.

Find the dimensions that yield the maximum amount of area for the pastures by completing the square.

THANKS for reading this and trying to help

**pickslides** · August 23rd 2009, 07:45 PM

Originally Posted by superfly8912

Three problems:

1) find the slope of the tangent line for the graph f(x)= x^2 at (2,4).

Find $f'(2)$

Spoiler:

**pickslides** · August 23rd 2009, 08:06 PM

Originally Posted by superfly8912

2) Find the slope of the line joining (2,4) and (2+h, f(2+h)) in terms of the nonzero number h. (This is also for the graph f(x)= x^2)

Let the slope of the line be m.

Find $m = \frac{f(2+h)-f(4)}{2+h-2}$

Spoiler:

Originally Posted by superfly8912

A rancher has 300 feet of fence to enclose two adjacent pastures. (See figure) Express the total area A of the two pastures as a function of x. What is the domain of A?

Graph the area function (I suppose I can do this once I find the equation) and estimate the dimensions that yield the maximum amount of area for the pastures.

Find the dimensions that yield the maximum amount of area for the pastures by completing the square.

Consider the perimeter of the fence.

$x+x+x+y+y+y+y = 300$

$3x+4y = 300$

Rearrange for y

$y = \frac{300-3x}{4}$

Now Area $= 2xy$ as there are 2 pastures

$A= 2xy = 2x\frac{300-3x}{4} =\frac{600x-6x^2}{4}$

Now graph this function. It is a quadratic so you can find the turning point (or the maximum) using symmetry.

**superfly8912** · August 24th 2009, 11:44 AM

Thank you for helping me out but I'm having trouble. Why did you square the numerator in number two? I happen to have the answer for this problem and apparently its 4 + h (but then again it could be a typo)...

**pickslides** · August 24th 2009, 01:54 PM

Originally Posted by superfly8912

2) Find the slope of the line joining (2,4) and (2+h, f(2+h)) in terms of the nonzero number h. (This is also for the graph f(x)= x^2)

$f(x)= x^2 \Rightarrow f(2+h) = (2+h)^2 = (2+h)(2+h) = 4+4h+h^2$

**Defunkt** · August 25th 2009, 01:21 AM

Originally Posted by pickslides

Let the slope of the line be m.

Find $m = \frac{f(2+h)-f(4)}{2+h-2}$

Should be:
$m = \frac{f(2+h)-f(2)}{2+h-2} = \frac{(2+h)^2 - (2)^2}{h} = \frac{h^2 + 4h + 4 - 4}{h} = h+4$

**pickslides** · August 25th 2009, 01:58 PM

Originally Posted by Defunkt

Should be:
$m = \frac{f(2+h)-f(2)}{2+h-2} = \frac{(2+h)^2 - (2)^2}{h} = \frac{h^2 + 4h + 4 - 4}{h} = h+4$

Yes I see... Nice work Defunkt!

Or even easier again

$m = \frac{f(2+h)-4}{2+h-2} = \frac{(2+h)^2 - 4}{h} = \frac{h^2 + 4h + 4 - 4}{h} = h+4$

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