How do you find the critical points for #f(x)=8x^3+2x^2-5x+3#?

Answer 1

Hello,

Determine the derivative.

#f'(x) = 24x^2 + 4x - 5#.
Solve #f'(x) = 0#. To do that, Calculate the discriminant #Delta = 4^2 - 4\times 24 times (-5) = 496#.
The critical points of #f# are the zeros of #f'#. So they are #(-4-sqrt(496))/48# and #(-4+sqrt(496))/48#.
You can simplify, because #496 = 16 \times 31# : #(-1-sqrt(31))/12# and #(-1+sqrt(31))/12#.
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 critical points of ( f(x) = 8x^3 + 2x^2 - 5x + 3 ), follow these steps:

  1. Compute the derivative of ( f(x) ) with respect to ( x ).
  2. Set the derivative equal to zero and solve for ( x ).
  3. The solutions obtained in step 2 are the critical points of the function.

Let's proceed with these steps:

  1. Compute the derivative of ( f(x) ): [ f'(x) = 24x^2 + 4x - 5 ]

  2. Set the derivative equal to zero and solve for ( x ): [ 24x^2 + 4x - 5 = 0 ]

This is a quadratic equation. You can use the quadratic formula to solve for ( x ): [ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ]

Where ( a = 24 ), ( b = 4 ), and ( c = -5 ). Substituting these values into the formula gives: [ x = \frac{{-4 \pm \sqrt{{4^2 - 4(24)(-5)}}}}{{2(24)}} ] [ x = \frac{{-4 \pm \sqrt{{16 + 480}}}}{{48}} ] [ x = \frac{{-4 \pm \sqrt{{496}}}}{{48}} ]

  1. Simplify and find the solutions for ( x ): [ x = \frac{{-4 \pm 4\sqrt{{31}}}}{{48}} ] [ x = \frac{{-1 \pm \sqrt{{31}}}}{{12}} ]

So, the critical points of the function ( f(x) = 8x^3 + 2x^2 - 5x + 3 ) are: [ x = \frac{{-1 + \sqrt{{31}}}}{{12}} ] [ x = \frac{{-1 - \sqrt{{31}}}}{{12}} ]

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