A cubic polynomial f is defined by: f(x)= 4x^3 + ax^2 + bx + k?

Question

a,b, and k are constants. There is a local maximum at x=-1, and the graph of f has a point of inlfection at x= -2. Find the values of a and b. If the integral f(x)dx from 0 to 1=32, find the value of k?

Answer 1

Step 1:
Differentiate, sub x = -1 and put this equal to 0 since there's a stationary point (local max) at x = -1,
i.e. put f'(-1) = 0
This will be an equation with unknowns a, b.
We need to find a second equation in order to solve for these.

Step 2:
Differentiate again, sub x = -2 and put this equal to 0, as there's an inflection at x = -2.
i.e. put f''( -2) = 0
This is our second equation with unknowns a, b. [Actually, b doesn't appear in this one, so it just gives us the value of a directly!]
Put it together with the one obtained in Step 1 and solve these simultaneous equations.

Step 3: Using the values a, b we have found, find a primitive for f(x) [this will still have the unknown k in it], sub x = 0, x = 1 and subtract. Put the result equal to 32, and solve for k.

I see now someone has already posted an answer outlining the same process. I don't think you need anyone else to do it?

If you want to check, I get
a = 24, b = 36, k = 5

Answer 2

The first derivative is 12x^2+2ax+b. For the local maximum to occur, 12-2a+b=0.
The second derivative is 24x+2a. If this inflects at x=-2, -48+2a=0 and a=24. If so, then b=36
We can integrate f(x) as
int ( 4x^3+ 24x^2+ 36x+k)= x^4+8x^3+18x^2+kx.
At 0, this integral is valued at 0
At 1, 1+8+18+k=32, and k=5

Answer 3

So we know that
f(1)' = 0
f(-2)" = 0

Thus f(x)' = 12x^2 + 2ax +b
Thus 0 = 12+2a+b ----- 2a+b= - 12.

Also f(x)" = 24x+2a --- a = -12.

Thus a = -12 and b = -36.

So f(x) = 4x^3 + 12x^2 - 36x + k.

We know the integral from 0 to 1 is 32.

The integral of f(x) = x^4 + 4x^3 - 18x + kx

If we plug in 1 we get 1 + 4 - 18 + k = 32.
Thus k = 45.

Thus f(x) = 4x^3 + 12x^2 - 36x + 45.

Answer 4

Start by differentiating the function

f'(x) = 12x^2 +2ax +b

From the question, we know that f'(x) = 0 at
x=-1 (local maximum) and
x=-2 (inflection)

These two equations should let you find values for a and b.

Then you go back to the original function and use actual figures for a and b.

Do the definite integral (you will get a value with a term in k). Compare your value with 32 to find the value of k.