Question

using the limit definition of the derivative find f'(x). Let f(x) = x^2 -4

Answer 1

f(x) = x^3 - 7x + 5
f(x + h) = (x + h)^3 - 7 * (x + h) + 5

f(x + h) - f(x) =>
(x + h)^3 - 7 * (x + h) + 5 - x^3 + 7x - 5 =>
x^3 + 3x^2 * h + 3x * h^2 + h^3 - 7x - 7h - x^3 + 7x =>
3x^2 * h + 3x * h^2 + h^3 - 7h

Divide by h

3x^2 + 3xh + h^2 - 7

Let h go to 0

3x^2 - 7

x = 5

3 * 5^2 - 7 =>
3 * 25 - 7 =>
75 - 7 =>
68

So we need a line with a slope of 68 that passes through (5 , 95)

y - 95 = 68 * (x - 5)
y - 95 = 68x - 340
y = 68x - 245 f(x) = x^3 - 7x + 5
f(x + h) = (x + h)^3 - 7 * (x + h) + 5

f(x + h) - f(x) =>
(x + h)^3 - 7 * (x + h) + 5 - x^3 + 7x - 5 =>
x^3 + 3x^2 * h + 3x * h^2 + h^3 - 7x - 7h - x^3 + 7x =>
3x^2 * h + 3x * h^2 + h^3 - 7h

Divide by h

3x^2 + 3xh + h^2 - 7

Let h go to 0

3x^2 - 7

x = 5

3 * 5^2 - 7 =>
3 * 25 - 7 =>
75 - 7 =>
68

So we need a line with a slope of 68 that passes through (5 , 95)

y - 95 = 68 * (x - 5)
y - 95 = 68x - 340
y = 68x - 245

Answer 2

what?

Answer 3

this your answer

Answer 4

medals now for me helping u

Answer 5

That's not the answer. I don't have a 7 anywhere in my problem.

Answer 6

@TheRealMeeeee stop copy/pasting random solutions that do not apply

Answer 7

im not so shut up

Answer 8

i wrote that myself

Answer 9

stupid lil boy

Answer 10

Well, you're wrong

Answer 11

who

Answer 12

http://openstudy.com/updates/529ce572e4b0b97c585d97da

Answer 13

def of derivative
\[\lim_{h \rightarrow 0} \frac{f(x+h) -f(x)}{h}\]
f(x) = x^2 -4
f(x+h) = (x+h)^2 -4 = x^2+2hx +h^2 -4
\[\lim_{h \rightarrow 0}\frac{(x^{2}+2xh+h^{2}-4) - (x^{2}-4)}{h} = \frac{2xh +h^{2}}{h}\]
now just factor out an "h" and see that it cancels with "h" on bottom , then apply limit

Answer 14

thank you!

Answer 15

yw later you will learn a shortcut for finding these derivatives and you wont have to use the long limit process :)