A cubic polynomial function f is defined by f(x) = 4x^3 + ax^2 + bx + k?
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.
a) Find the values of a and b
b) If ∫ (from 0 to 1) f(x) dx = 32, what is the value of k?
I found a=24 and b=36
i don't know about B
To find the values of a and b, we can use the information given about the local minimum and point of inflection.
a) Since the function f has a local minimum at x = -1, this means that the derivative of f with respect to x is equal to zero at x = -1. Taking the derivative of f(x), we get:
f'(x) = 12x^2 + 2ax + b
Setting f'(x) equal to zero:
12(-1)^2 + 2a(-1) + b = 0
12 - 2a + b = 0
b = 2a - 12
Now, since the graph of f has a point of inflection at x = -2, the second derivative of f with respect to x must be zero at x = -2. Taking the second derivative of f(x), we get:
f''(x) = 24x + 2a
Setting f''(x) equal to zero:
24(-2) + 2a = 0
-48 + 2a = 0
2a = 48
a = 24
Substituting the value of a into the equation for b:
b = 2(24) - 12
b = 48 - 12
b = 36
So the values of a and b are a = 24 and b = 36.
b) To find the value of k, we can use the given integral ∫ (from 0 to 1) f(x) dx = 32. We'll integrate the function from 0 to 1 and set it equal to 32:
∫ (from 0 to 1) (4x^3 + ax^2 + bx + k) dx = 32
Integrating term by term:
(4/4)x^4 + (a/3)x^3 + (b/2)x^2 + kx (from 0 to 1) = 32
Simplifying further:
x^4 + (a/3)x^3 + (b/2)x^2 + kx (from 0 to 1) = 32
Now, let's substitute the limits of integration (0 and 1) into the equation and simplify:
(1^4 + a/3(1^3) + b/2(1^2) + k(1)) - (0^4 + a/3(0^3) + b/2(0^2) + k(0)) = 32
(1 + a/3 + b/2 + k) - (0 + 0 + 0 + 0) = 32
1 + a/3 + b/2 + k = 32
Rearranging the equation:
a/3 + b/2 + k = 31
Now, substitute the known values of a and b we found earlier into the equation:
24/3 + 36/2 + k = 31
8 + 18 + k = 31
26 + k = 31
k = 31 - 26
k = 5
Therefore, the value of k is 5.