1) Let f(x)= x^3-10x calculate the difference quotient f(2+h)-f(2)/h for h= .1 _

Question

1)

Let f(x)= x^3-10x calculate the difference quotient f(2+h)-f(2)/h for

h= .1 __________ ??

h= .01 _________??

h= -.01 _________??

h= -.1 _________??

If someone now told you the derivative (slope of the tangent line to the graph) of f(x) at x=2 was an integer, what would you expect it to be? _______??

2)

Let f(x) = 1/(x-8) calculate the difference quotient f(5+h)-f(5)/h for

h= .1 ______??

h= .01 _______??

h= -.01 ________??

h= -.1 ______??

If someone now told you the derivative (slope of the tangent line to the graph) of f(x) at x=5 was (-1/n^2) for some integer n what would you expect n to be? ______??

PLEASE ANSWER ALL PARTS OF THE QUESTION!!!!!!!!

Explanation / Answer

Let f(x)= x^3-10x calculate the difference quotient f(2+h)-f(2)/h for

h= .1 : 2.61

h= .01 : 2.061

h= -.01 : 1.9401

h= -.1 : 1.41

If someone now told you the derivative (slope of the tangent line to the graph) of f(x) at x=2 was an integer, what would you expect it to be?

Answer: 2
Because for h = 0.01, its close to 2 and f'(2) = lim(h->0) [f(2+h)-f(2)/h]

2)

Let f(x) = 1/(x-8) calculate the difference quotient f(5+h)-f(5)/h for

h= .1 : ?0.114942529

h= .01 : ?0.11148272

h= -.01 : ?0.110741971

h= -.1 : ?0.107526882

If someone now told you the derivative (slope of the tangent line to the graph) of f(x) at x=5 was (-1/n^2) for some integer n what would you expect n to be?

Answer: n=3
Because for h = 0.01, its close to 1/f'(5) is close to 9 and f'(5) = lim(h->0) [f(5+h)-f(5)/h]

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