Curve Sketching: Solving for b and d: f(x)=x^3+bx^2+d

  • #1

Homework Statement



f(x)=x^3+bx^2+d given a critical point of (2,-4), solve for b and d

Homework Equations



f'(x)=3x^2+2bx ?

The Attempt at a Solution



I wasn't really sure where to go with this.

So I know that a critical point happens when f'(x)=0, and this point can be either a max, min, or extrema.

I tried making the derivative = to 0 and then subbed in the x value from the critical point of x=2. But i believe that this is definitely not right.

Any help is much appreciated, thank you.
 

Answers and Replies

  • #2
33,722
5,418

Homework Statement



f(x)=x^3+bx^2+d given a critical point of (2,-4), solve for b and d

Homework Equations



f'(x)=3x^2+2bx ?

The Attempt at a Solution



I wasn't really sure where to go with this.

So I know that a critical point happens when f'(x)=0, and this point can be either a max, min, or extrema.

I tried making the derivative = to 0 and then subbed in the x value from the critical point of x=2. But i believe that this is definitely not right.

Any help is much appreciated, thank you.
From the given information, you know that f(2) = 4 and f'(2) = 0. This will give you two equations in your two unknowns so that you can solve for b and d.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
956

Homework Statement



f(x)=x^3+bx^2+d given a critical point of (2,-4), solve for b and d

Homework Equations



f'(x)=3x^2+2bx ?

The Attempt at a Solution



I wasn't really sure where to go with this.

So I know that a critical point happens when f'(x)=0, and this point can be either a max, min, or extrema.

I tried making the derivative = to 0 and then subbed in the x value from the critical point of x=2. But i believe that this is definitely not right.

Any help is much appreciated, thank you.
No, that definitely is right! So what did you get when you put x= 2 into f'(x)= 0?
 
  • #4
Oh yeeesss! Thank you! I knew i would have to get two unknowns from somewhere then solve!
 

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