# AP Calculus Question

A cubic polynomial function f is defined by:
f(x)=4x^3+ax^2+bx+k
where a, b, and k are constants. The function f has a local minimum at x=-1, and the graph of f has a point of inflection at x=-2.

1) Find the values of a and b
2) If the integral from 0 to 1 of f(x)dx = 32, what is the value of k?

1. f(x)=4x^3+ax^2+bx+k

2. Therefore, f'(x)=12x^2+2ax+b, f”(x)=24x+2a

3. Solve: f'(-1)=12-2a+b=0, f”(-2)=-48+2a=0

4. Obtaining: a=24, b=36

5. The integral from 0 to 1 of f(x)dx = 32 -> 27+k=32 -> k=5

### 9 Responses to “AP Calculus Question”

1. Luke says:

Wow…I remember how to do that and it’s been 3 years since I’ve really worked with that sort of Calc.

Fun, fun, fun.

2. Erica says:

I did this so easy.
I guess being in grade 8 and doing Calc. really paid off.
That was fun!

3. Olivia says:

The both of you are really arrogant. Thanks, we’re all completely disgusted, annoyed, and unimpressed (especially since Erica apparently can’t use proper grammar). But as someone that wants to leave a real message, thanks for the help Elliot. Your explanation was really clear and easy to understand!

4. Sam says:

I don’t understand where you are getting the 27 from in answer number 5?

5. a dictionary says:

no one in the world would understand what you mean

only if they already understood what a polynomial, inflections, constants, intergral, local minimum, your varialbles what they mean, what you mean and how they work in the real world.

if you were writing an essay on calculus and used this you’d get zero marks.

in english you have to make sense.

6. a friend of mine was trying to find answer here but all she got is bs.

you guys are gay.

hope you don’t get a F in the classes. suckers

7. Kev says:

To clarify question number 5:

– Now that you have the values of ‘a’ and ‘b’, plug them into the original cubic polynomial.
[4x^3 + 24x^2 + 36x + k]

– Since the problem states the integral from 0 to 1 of f(x)dx = 32, you need to integrate cubic polynomial.
[4x^3 + 24x^2 + 36x + k] –> [x^4 + 8x^3 + 18x^2 + kx]

– Now plug in the values of ‘0’ and ‘1’. Simplify.
[(1)^4 + 8(1)^3 + 18(1)^2+ k(1)] = [k + 27]
[(0)^4 + 8(0)^3 + 18(0)^2+ k(0)] = 

– You must not solve for the integral. When plugging in ‘0’ for ‘x’, the value came out as ‘0’. So you can disregard it.
[k + 27] =  –> [k=5]

8. Kev says:

SORRY I FIXED THE GRAMMER

To clarify question number 5:

– Now that you have the values of �a� and �b�, plug them into the original cubic polynomial.
[4x^3 + 24x^2 + 36x + k]

– Since the problem states the integral from 0 to 1 of f(x)dx = 32, you need to integrate the cubic polynomial.
[4x^3 + 24x^2 + 36x + k] �> [x^4 + 8x^3 + 18x^2 + kx]

– Now plug in the values of �0? and �1?. Simplify.
[(1)^4 + 8(1)^3 + 18(1)^2+ k(1)] = [k + 27]
[(0)^4 + 8(0)^3 + 18(0)^2+ k(0)] = 

– You must now solve for the integral. When plugging in �0? for �x�, the value came out as �0?. So you can disregard it. Therefore:
[k + 27] =  �> [k=5]

9. Matthew Slaby says:

Please show me how to integrate a cubic polynomial to the -0.5 power.
I suppose that I could factor it into a binomial and a quadratic and then go from there.
I was hoping that the Table of Integrals the I have is incomplete. Therefore, any assitance is welcome. Don’t worry if this is out of your expertise.

Kind Regards,
Matt