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

[Buzzlastyear]

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.

 

[Mark44]

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.

 

[HallsofIvy]

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?

 

[Buzzlastyear]

Oh yeeesss! Thank you! I knew i would have to get two unknowns from somewhere then solve!