Question

Find all the values of x for which the function is differentiable.

f(x)= (x^3-8)/(x^2-4x-5)

Answer 1

To find the values of x for which the function f(x) = (x^3-8)/(x^2-4x-5) is differentiable, check for any discontinuities or vertical asymptotes in the function. A function is differentiable if it is continuous and has a defined derivative at that point.

Step-by-step explanation:

To find the values of x for which the function f(x) = (x^3-8)/(x^2-4x-5) is differentiable, we need to check for any discontinuities or vertical asymptotes in the function. A function is differentiable at a point if it is continuous and has a defined derivative at that point.

One way to check for discontinuities is to look for values of x that make the denominator of the function equal to zero. So, we set x^2-4x-5 = 0 and solve for x. The roots of this quadratic equation are x = -1 and x = 5.

Therefore, the function f(x) is differentiable for all values of x except x = -1 and x = 5.

Answer 2

The function is differentiable for:
x² - 4 x - 5 ≠ 0
x² - 5 x + x - 5 ≠ 0
x ( x - 5 ) + ( x - 5 ) ≠ 0
( x - 5 ) ( x + 1 ) ≠ 0
x ≠ 5 and x ≠ - 1
Answer: x ∈ R \ { -1, 5 }