Respuesta :
Answer:
x = -6 and x = 4
Step-by-step explanation:
In math, the critical points of a function are the points where the derivative equals zero.
So, first we will find the derivative of the function. The derivative is:
[tex]f'(x)=3x^2 +6x-72[/tex]
Now, we are going to make the derivative equal zero and find the answers of the equation.
[tex]3x^2 +6x-72=0\\3(x^2 +2x-24)=0\\3(x+6)(x-4)=0\\[/tex]
So we have that the critical points are the answers to this equation:
[tex]x+6= 0 \\x= - 6[/tex]
and
[tex]x-4=0\\x=4[/tex]
Thus, the critical points are x=-6 and x=4
Using it's concept, it is found that the critical numbers of the function are:
x = -6 and x = 4.
The critical numbers of a function f(x) are the values of x for which it's derivative is zero, that is:
[tex]f^{\prime}(x) = 0[/tex]
In this problem, the function is:
[tex]f(x) = x^3 + 3x^2 - 72x[/tex]
The derivative is:
[tex]f^{\prime}(x) = 3x^2 + 6x - 72[/tex]
[tex]f^{\prime}(x) = 3(x^2 + 2x - 24)[/tex]
Then:
[tex]f^{\prime}(x) = 0[/tex]
[tex]3(x^2 + 2x - 24) = 0[/tex]
[tex]x^2 + 2x - 24 = 0[/tex]
Which is a quadratic equation with coefficients [tex]a = 1, b = 2, c = -24[/tex], which we have to solve.
[tex]\Delta = 2^2 - 4(1)(-24) = 100[/tex]
[tex]x_{1} = \frac{-2 + \sqrt{100}}{2} = 4[/tex]
[tex]x_{2} = \frac{-2 - \sqrt{100}}{2} = -6[/tex]
The critical numbers of the function are x = 4 and x = -6.
A similar problem is given at https://brainly.com/question/16944025
