Question

For what values of x is the function f(x) = 4x²-6x/4x³-2x²-6x differentiable?

O all real numbers except -1
O all real numbers except -1 and 0
O all real numbers except -1.0, and 3/2
O all real numbers except -1,2 and 3/2
O all real numbers

Answer 1

The function is not differentiable at
`\( x = 0, (3)/(2), -1 \)`. The correct option is: (C) all real numbers except -1, 0, and 3/2

How to find non-differentiable function?

To determine where the function
`\( f(x) = (4x^2 - 6x)/(4x^3 - 2x^2 - 6x) \)` is differentiable:

The function is defined for all real numbers except where the denominator equals zero. Therefore, we need to find the values of x that make the denominator zero:

`\[ 4x^3 - 2x^2 - 6x = 0 \]`

Factoring out 2x:

`\[ 2x(2x^2 - x - 3) = 0 \]`

Now, solve for x:

`\[ 2x = 0 \implies x = 0 \]`

`\[ 2x^2 - x - 3 = 0 \]`

Using the quadratic formula, find the roots:

`\[ x = (1 \pm √(25))/(4) \]`

So, the values of x that make the denominator zero are
`\( x = 0, (3)/(2), -1 \)`.

Therefore, the function is not differentiable at
`\( x = 0, (3)/(2), -1 \)`. The correct option is:

`\[ \boxed{\text{O all real numbers except } -1, (3)/(2), \text{ and } 0} \]`