Find the value(s) of x, so that f’(x) = 0 for the function f(x) = 4x3 – 6x2 – 24x.
I will assume you meant
f(x) = 4x^3 – 6x^2 – 24x
then
f'(x) = 12x^2-12x - 24
= 0
divide by 12
x^2 - x - 2 = 0
(x-2)(x+1) = 0
x = 2 or x = -1
Oh, finding the value(s) of x where f'(x) = 0? That's a job for the math police! We need to take the derivative of f(x) = 4x^3 – 6x^2 – 24x.
So, let's put on our detective hats and differentiate this crime scene! f'(x) = 12x^2 – 12x – 24.
Now we are on the hunt for those sneaky zeros! Set f'(x) = 0 to crack this mystery. 12x^2 – 12x – 24 = 0.
To make this a little more manageable, let's divide everything by 12. x^2 – x – 2 = 0.
Ah, a classic case of a quadratic equation! We need some x-values that will satisfy this quadratic equation and bring justice to f'(x).
Using our secret weapon, the quadratic formula, we find x = (-b ± √(b^2 - 4ac))/(2a). Plugging in the values, we get x = (1 ± √(1^2 - 4(1)(-2)))/(2(1)).
Calculating further, we have x = (1 ± √(1 + 8))/2.
Looks like we have two suspects! Solving further, we find x = (1 ± √9)/2.
After interrogating these suspects, it turns out that x = (1 ± 3)/2.
So, the value(s) of x that make f'(x) = 0 are x = -1 and x = 2. Case closed!
To find the value(s) of x where f'(x) = 0 for the function f(x) = 4x^3 - 6x^2 - 24x, we need to find the critical points.
The derivative of f(x) represents the rate of change of f(x) with respect to x, so we need to find f'(x) and set it equal to 0.
Step 1: Find f'(x)
To find the derivative of f(x), we differentiate each term with respect to x using the power rule:
f(x) = 4x^3 - 6x^2 - 24x
f'(x) = 12x^2 - 12x - 24
Step 2: Set f'(x) = 0 and solve for x
12x^2 - 12x - 24 = 0
Step 3: Simplify the equation
Divide the equation by 12 to simplify it:
x^2 - x - 2 = 0
Step 4: Factor the equation
We can factor the equation x^2 - x - 2 = 0 by finding two numbers whose product is -2 and whose sum is -1. The factors are -2 and 1, so we have:
(x - 2)(x + 1) = 0
Setting each factor equal to 0, we get:
x - 2 = 0 or x + 1 = 0
Solving for x, we have:
x = 2 or x = -1
Therefore, the value(s) of x where f'(x) = 0 for the function f(x) = 4x^3 - 6x^2 - 24x are x = 2 and x = -1.
To find the value(s) of x for which f'(x) = 0, we need to find the critical points of the function f(x). The critical points occur where the derivative of the function is equal to zero or undefined.
First, let's find the derivative f'(x) of the function f(x) = 4x^3 - 6x^2 - 24x:
f'(x) = 12x^2 - 12x - 24.
Next, we'll set f'(x) = 0 and solve for x:
12x^2 - 12x - 24 = 0.
Dividing the equation by 12 to simplify, we get:
x^2 - x - 2 = 0.
Now, we can factor the quadratic equation:
(x - 2)(x + 1) = 0.
Setting each factor equal to zero, we have:
x - 2 = 0 or x + 1 = 0.
Solving each equation, we find:
x = 2 or x = -1.
Therefore, the values of x that make f'(x) = 0 are x = 2 and x = -1.