Fnd the values of x where the function y=x^3-3x^2+2 reaches a maximum and minimum, values of the function. Sketch the function.Fnd the values of x where the function y=x^3-3x^2+2 reaches a maximum...

Fnd the values of x where the function y=x^3-3x^2+2 reaches a maximum and minimum, values of the function. Sketch the function.

Fnd the values of x where the function y=x^3-3x^2+2 reaches a maximum and minimum, values of the function. Sketch the function.

Asked on by lavo

5 Answers | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should know that the function rea = 0ches its maximum or minimum at values of x that represent the zeroes of f'(x).

You need to solve for x the equation f'(x) = 0 such that:

(x^3-3x^2+2)' = 0 => 3x^2 - 6x = 0 => x^2 - 2x = 0

Factoring out x yields:

x(x-2) = 0 => x = 0 and x-2 = 0 => x = 2

You should know that the expression x^2 - 2x is negative if x in (0,2) and it is positive if x in (-oo,0) or (2,oo).

Hence, the function reaches its maximum at x = 0 and it reaches its minimum at x = 2 as the graph below shows:

albimaia's profile pic

albimaia | Student, College Freshman | (Level 1) Honors

Posted on

I'm confused. In the first answer, for x=2, you put

y2 = 3^3 - 3*2^2 + 2=17

Why?

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The local extreme points of a function could be calculated using the first derivative test.

We know that the values where first derivative is cancelling are representing the values where the function has an extreme point.

To find out the roots of the derivative of the function, first, we'll have to calculate it:

f'(x) = y' = (x^3-3x^2+2)'

f'(x) = (x^3)' - (3x^2)' + (2)'

f'(x) = 3x^2 - 3*2x + 0

f'(x) = 3x^2 - 6x

We'll set the first derivative as zero:

3x^2 - 6x = 0

We'll factorize by 3x:

3x*(x-2) = 0

We'll set each factor from the product as zero:

3x = 0

We'll divide by 3:

x = 0

x-2 = 0

We'll add 2:

x = 2

The first derivative is negative between the roots and it's positive outside the roots.

That means that the function is increasing over the interval (-inf.,0) and it's decreasing over the interval (0,2). That means that for x = 0, the function has a maximum.

f(0) = (0^3-3*0^2+2)

f(0) = 2

The maximum point has the coordinates: (0 , 2).

Also, the function is decreasing over the interval(0,2) and it's increasing over the interval (2, +inf.). That means that for x = 2, the function has a minimum.

f(2) = (2^3-3*2^2+2)

f(2) = 8 - 12 + 2

f(2) = -2

The minimum point has the coordinates: (2 , -2).

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

y = x^3-3x^2+2 .

To find the maximum and muninum.

Solution: We know that f(x) reaches  maximum or minimum for ththe value x =c of for which f'(c) = 0 and f''(c) < 0  or f"(c) => 0.

So we find f'x)  and set f'(x ) = 0, and the solutioo x= c. And see whether f"(c) <0  for maximum Or f"(c) > 0 for minimum.

y' = (x^3-3x^2+2 )' = 0

3x^2-6x = 0

3x(x-2). So x= 0 or x= 2  are the values of c.

f''(x) =6x - 6. 

So f''(0) = 6*0 -3 = -3 < 0 . So f(0) = 0^3-3*0+2 = 2 is maximum.

f"(2) = 6*2-6 = 6 > 0. f(2) is the minimum.

f''(2) = 2^3-3*2^2+2 = 8-12+2 = -2 is the minimum for x = 2.

 

Since f(1) = 0, (x-1) is factor. So we can write,

x^3-3x^2+2 = (x-1)(x^2+kx -2)

Coefficients of x^3 and x^3 and  conts are equal on both sides.

So equating the  x 's on both sides: 0x = -kx-2x = 0, k = -2.

Equating x^2' s on both sides: -3x^2 = kx^2-x^2, or -2x^2 = kx^2. So k = -2.

So x^2-3x^2+2 = (x-1)(x^2-2x-2).

x^2-2x-2 has roots ,

x1 = {- -2+sqrt(4+8)}/2 = 1+sqrt3, Or

x2 = 1-sqrt3.

Therefore

f(x) = x^3 -3x^2+2 has 3 real roots x = 1, x1 = 1+sqrt3 and x2 = 1- sqrt3.

So f(x) crosses x axis at x = 1- sqrt3, x= 1 and x = 1+sqrt 3.

Since f'(x) =  3x^2-6x = 3x(x-2) is positive for f(x) is an increasing function for x >2 and for x <0.

f('x)  = 3x(x-2) is > for x belonging to the interval (0, 2).

Therefore,

for x > 2, f(x) is incresing and f(x) goes infinite as x--> infinity.

 0 < x <2 , f(x) is decreasing function.

f(x) = 0. Or f(x) crosses x axis at x=  1+sqrt3, 1 and 1-sqrt3.

At  x = 0, f(x) crosses y axis at y = 2.

For x < 0,  f(x) is an increasing finction again . f(x) approach - infinity ax x--> -infinity.

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

When the value of function y is maximum or minimum the value of dy/dx is equal to 0.

Given:

y = x^3 - 3x^2 + 2

Therefore:

dy/dx = 3x^2 - 3*2x = 3x(x - 2)

Equating above expression for dy/dx to 0, we get two values of x.

x 1 = 0 and x2 = 2

To get corresponding value of function, we substitute these values of y in equation for y. Thus:

y1 = 0^3 - 3*0^2 + 2

= 2

y2 = 3^3 - 3*2^2 + 2

= 27 - 12 + 2

= 17

Answer:

Maximum = 17

Minimum = 2

We’ve answered 318,955 questions. We can answer yours, too.

Ask a question