How do you find the local max and min for # f(x)=x^3-2x+5# on the interval (-2,2)?

Answer 1

See below.

First find the 1st derivative of #x^3-2x+5#
#d/dx(x^3-2x+5)=3x^2-2#
Maximum and minimum points have a horizontal gradient .i.e. gradient of #0#. Solving the 1st derivative will allow us ti identify these points.
#f'(x)=0#
#:.#
#3x^2-2=0=>x=sqrt(6)/3 , x=-sqrt(6)/3#

Plugging these values into the second derivative, will allow us to find whether these are maximum, minimum or points of inflection. Using the following, if:

#f''(x)>0color(white)(8888)# minimum value
#f''(x)<0color(white)(8888)# maximum value
#f''(x)=0color(white)(8888)# maximum/minimum or point of inflection.

The second derivative is the derivative of the first derivative.

#:.#
#d/dx(3x^2-2)=6x#
Plugging in our values of #x#:
#6(sqrt(6)/3)=2sqrt(6)color(white)(8888888)# ( minimum value )
#6(-sqrt(6)/3)=-2sqrt(6)color(white)(88)# ( maximum value )
Both #color(white)(8888)x=sqrt(6)/3 , x=-sqrt(6)/3# are in #( -2 ,2)#

GRAPH:

graph{y=x^3-2x+5 [-5, 5, -20, 20]}

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the local maximum and minimum of ( f(x) = x^3 - 2x + 5 ) on the interval ((-2, 2)), follow these steps:

  1. Compute the derivative of ( f(x) ) to find critical points. [ f'(x) = 3x^2 - 2 ]

  2. Set ( f'(x) = 0 ) and solve for ( x ) to find critical points. [ 3x^2 - 2 = 0 ] [ x^2 = \frac{2}{3} ] [ x = \pm \sqrt{\frac{2}{3}} ]

  3. Evaluate ( f(x) ) at the critical points and endpoints of the interval ((-2, 2)). [ f\left(-\sqrt{\frac{2}{3}}\right), \quad f\left(\sqrt{\frac{2}{3}}\right), \quad f(-2), \quad f(2) ]

  4. Compare the values of ( f(x) ) at critical points and endpoints to determine local maxima and minima.

After evaluating, you'll find:

[ f\left(-\sqrt{\frac{2}{3}}\right) \approx 6.50 ] [ f\left(\sqrt{\frac{2}{3}}\right) \approx 3.50 ] [ f(-2) = 13 ] [ f(2) = 5 ]

The minimum occurs at ( x = -\sqrt{\frac{2}{3}} ) and the maximum occurs at ( x = 2 ).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7