Question

Find all values of x for which f(x)=(x^3-8)/(x^2-4x-5) is differentiable. Be sure to explain/justify your answer. Include where f'(x) does not exist and state why.

Answer 1

well for mine the graph has 2 vertical asymptotes and an oblique asymptote...

the vertical asymptotes will identify the restrictions in the domain...

Answer 2

so what does that mean?

Answer 3

it means that there are gaps in the graph where the curve doesn't exist...

Answer 4

well you are asked to identify where the curve is differentiable... so if a curve has a gap in the domain, you can't find a derivative at that point...

so you can differentiate using the quotient rule...

then identify the restrictions in the denominator of the derivative

Answer 5

@campbell_st For the derivative I got (x^4-8x^3-15x^2+16x-32)/((x^2-4x-5)^2).

Answer 6

ok... so the derivative doesn't exist for the values of x that make the denominator zero

so the denominator can be factored to

((x -5)(x +1))^2

so you need to find the values of x that make the denominator zero..

Answer 7

ok I got x=-1 and x=5

Answer 8

ok... so you now know where f'(x) doesn't exist

and the same values of x, in the original function, cause gaps in the curve...

these are known as vertical asymptotes

here is a graph of the curve so you can see the vertical asymptotes and why they cause gaps

hope it helps

Answer 9

so how do I find the values x where the function is differentiable?

Answer 10

well you use inequalities

as an example is differentiable for -1 < x < 5

also

\[x < -1.... and .... x > 5\]

Answer 11

or you could say

all real x except x = -1 and x = 5

Answer 12

I hope it all helps...

Answer 13

so wait the answer to the values of x is all real numbers except x = -1 and x = 5? its the same as the ones that don't exist?

Answer 14

thats it... its all to do with the denominator...

differentiating only raises the power of the denominator...

have to go...

good luck

Answer 15

thanks!