Find (and classify) the critical points of the following function and determine if they are local max, local min, or neither: f(x) =2x^3 + 3x^2 - 120x

Question

25-10-2022
Mathematics

contestada

Find (and classify) the critical points of the following function and determine if they are local max, local min, or neither: f(x) =2x^3 + 3x^2 - 120x

Respuesta :

Otras preguntas

The branch of public law that deals with the various regulatory functions and activities of agencies of the government is: a. administrative law. b. substantive

Kiran creates a scatter plot showing the relationship between the number of students attending drama club and the number of students attending poetry club each

A pie has a circumference of 31.4 in. What is the diameter of the pie?

A home goods store purchased a desk lamp and marked it up 155% from the original cost of $9.12. Then, wanting to make room for summer inventory, the store place

What is 48/54 in its lowest form?

HELP ME PLEASE !! In 3-5 sentences, Explain 2 problems with the evolutionary belief of long stable periods of no mutations and short periods of rapid mutations.

What are the square number factors of 56?

Please arrange each event in chronological order, from earliest to latest Question 4 options: Thomas Jefferson won the Election of 1800, becoming the first non-

Which of the following is true? Select one: O a. 212°F = 100°C O b. All of the above O c. -460°F = -273°C O d. 32°F = 0°C

An event in which the heart abruptly stops beating or develops an arrhythmia that prevents it from pumping blood effectively is called.

MoziahN556205 MoziahN556205 · Answer 1 · 2022-10-25T17:29:38+02:00

As given by the question

There are given that the function:

[tex]f(x)=2x^3+3x^2-120x[/tex]

Now,

To find the critical point, differentiate the given function with respect to x and put the result of function equal to zero

So,

[tex]\begin{gathered} f(x)=2x^3+3x^2-120x \\ f^{\prime}(x)=6x^2+6x-120 \end{gathered}[/tex]

Then,

[tex]\begin{gathered} f^{\prime}(x)=0 \\ 6x^2+6x-120=0 \\ x^2+x-20=0 \\ x^2+5x-4x-20=0 \\ x(x+5)-4(x+5) \\ (x-4)(x+5) \\ x=4,\text{ -5} \end{gathered}[/tex]

Now,

To find the y-coordinate, we need to substitute the above value, x = 4, -5, into the function f(x)

So,

First put x = 4 into the given function:

[tex]\begin{gathered} f(x)=2x^3+3x^2-120x \\ f(4)=2(4)^3+3(4)^2-120(4) \\ =128+48-480 \\ =-304 \end{gathered}[/tex]

And,

Put x = -5 into the function f(x):

[tex]\begin{gathered} f(x)=2x^3+3x^2-120x \\ f(-5)=2(-5)^3+3(-5)^2-120(-5) \\ =-250+75+600 \\ =425 \end{gathered}[/tex]

Hence, the critical point is, (4, -304) and (-5, 425).

Now,

To find the local maxima and local minima, we need to find the second derivative of the given function:;

So,

[tex]\begin{gathered} f^{\prime}(x)=6x^2+6x-120 \\ f^{\doubleprime}(x)=12x+6 \end{gathered}[/tex]

Now,

The put the value from critical point into the above function to check whether it is maxima or minima.

So,

First put x = 4 into above function:

[tex]\begin{gathered} f^{\doubleprime}(x)=12x+6 \\ f^{\doubleprime}(4)=12(4)+6 \\ f^{\doubleprime}(4)=48+6 \\ f^{\doubleprime}(4)=54 \\ f^{\doubleprime}(4)>0 \end{gathered}[/tex]

And,

Put x = -5 into the above function

[tex]\begin{gathered} f^{\doubleprime}(x)=12x+6 \\ f^{\doubleprime}(-5)=12(-5)+6 \\ f^{\doubleprime}(-5)=-60+6 \\ f^{\doubleprime}(-5)=-54 \\ f^{\doubleprime}(-5)<0 \end{gathered}[/tex]

Then,

According to the concept, if f''(x)>0 then it is local minima function and if f''(x)<0, then it is local maxima function

Hence, the given function is local maxima at (-5, 425) and the value is -54 and the given function is local minima at point (4, -304) and the value is 54.