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)
From a=24,b=36,
we find
f'(x)=0 has roots at x=-3 and -1, so that checks.
f"(x)=0 has a root at x=-2, so that checks too.
To solve B, we integrate
f(x):=4*x^3+24*x^2+36*x+k
from 0 to 1:
I=∫f(x)dx
=x^4+8*x^3+18*x^2+k*x
Evaluate I from 0 to 1 to give k+27
But since is given as I=32, so
I=32=k+27
Solve for k.
Well, well, well! It seems you've already made some progress solving part (a) of the problem. Kudos to you!
Now, let's tackle part (b). We are given that ∫ (from 0 to 1) f(x) dx = 32. To find the value of k, we can integrate the given function from 0 to 1 and set it equal to 32.
∫ (from 0 to 1) f(x) dx = 32
∫ (from 0 to 1) (4x^3 + ax^2 + bx + k) dx = 32
To simplify things, let's integrate each term separately:
∫ (from 0 to 1) 4x^3 dx + ∫ (from 0 to 1) ax^2 dx + ∫ (from 0 to 1) bx dx + ∫ (from 0 to 1) k dx = 32
Integrating each term gives us:
[ x^4 ] (from 0 to 1) + [ (a/3)x^3 ] (from 0 to 1) + [ (b/2)x^2 ] (from 0 to 1) + [ kx ] (from 0 to 1) = 32
Now, let's plug in the values:
(1^4 - 0^4) + (a/3)(1^3 - 0^3) + (b/2)(1^2 - 0^2) + k(1 - 0) = 32
1 + a/3 + b/2 + k = 32
We can simplify further:
a/3 + b/2 + k = 31
Now, since we already found the values for a and b as a = 24 and b = 36, we can substitute them in:
24/3 + 36/2 + k = 31
8 + 18 + k = 31
26 + k = 31
k = 31 - 26
k = 5
So, the value of k is 5.
I hope this solves the mystery of part (b) for you. Keep up the good work!
To solve for the values of a and b, we can use the given information about the local minimum and point of inflection.
a) Local Minimum:
The function f has a local minimum at x = -1 when its derivative is equal to zero.
Let's find the derivative of f(x) first:
f'(x) = 12x^2 + 2ax + b
Setting f'(x) equal to zero:
12(-1)^2 + 2a(-1) + b = 0
12 - 2a + b = 0 (Equation 1)
b) Point of Inflection:
The graph of f has a point of inflection at x = -2 when its second derivative changes sign.
Let's find the second derivative of f(x):
f''(x) = 24x + 2a
To find the sign change in the second derivative at x = -2, substitute -2 into f''(x):
24(-2) + 2a < 0
-48 + 2a < 0
2a < 48
a < 24 (Equation 2)
c) Solving Equations 1 and 2 simultaneously:
From Equation 2, we know that a must be less than 24.
Substituting the value of a from Equation 1 into Equation 2:
12 - 2a + b = 0
12 - 2(24) + b = 0
12 - 48 + b = 0
b - 36 = 0
b = 36
Therefore, a = 24 and b = 36.
b) To find the value of k, we can use the given definite integral:
∫ (from 0 to 1) f(x) dx = 32
Substituting the function f(x) into the integral:
∫ (from 0 to 1) (4x^3 + ax^2 + bx + k) dx = 32
Integrating term by term:
∫ (from 0 to 1) 4x^3 dx + ∫ (from 0 to 1) ax^2 dx + ∫ (from 0 to 1) bx dx + ∫ (from 0 to 1) k dx = 32
(1/4)x^4 + (1/3)ax^3 + (1/2)bx^2 + kx ∣(from 0 to 1) = 32
[(1/4)(1)^4 + (1/3)a(1)^3 + (1/2)b(1)^2 + k(1)] - [(1/4)(0)^4 + (1/3)a(0)^3 + (1/2)b(0)^2 + k(0)] = 32
(1/4) + (1/3)a + (1/2)b + k = 32
Simplifying:
1/4 + (1/3)(24) + (1/2)(36) + k = 32
1/4 + 8 + 18 + k = 32
1/4 + 26 + k = 32
k = 32 - 26 - 1/4
k = 5 + 3/4
k = 20/4 + 3/4
k = 23/4
Therefore, the value of k is 23/4.
To find the values of a and b in the cubic polynomial function f(x) = 4x^3 + ax^2 + bx + k, we can use the information given about the local minimum and point of inflection.
a) Local Minimum:
A local minimum occurs at a point where the derivative of the function equals zero. In this case, the local minimum occurs at x = -1, so we can find the derivative of the function and set it equal to zero to solve for a.
1. Differentiate the function f(x) = 4x^3 + ax^2 + bx + k with respect to x:
f'(x) = 12x^2 + 2ax + b
2. Set f'(x) equal to zero and substitute x = -1:
12(-1)^2 + 2a(-1) + b = 0
12 + (-2a) + b = 0
3. Simplify the equation:
-2a + b = -12 (equation 1)
b) Point of Inflection:
A point of inflection occurs where the second derivative of the function equals zero. In this case, the point of inflection occurs at x = -2, so we can find the second derivative of the function and set it equal to zero to solve for a.
1. Differentiate the derivative f'(x) with respect to x:
f''(x) = 24x + 2a
2. Set f''(x) equal to zero and substitute x = -2:
24(-2) + 2a = 0
-48 + 2a = 0
3. Simplify the equation:
2a = 48
a = 24 (equation 2)
Now that we have found the value of a, we can substitute it back into equation 1 to solve for b.
-2(24) + b = -12
-48 + b = -12
b = -12 + 48
b = 36
So, the values of a and b are a = 24 and b = 36.
b) To find the value of k, we are given the definite integral of the function from x = 0 to x = 1, which means we need to find the area under the curve between these limits.
∫ (from 0 to 1) f(x) dx = 32
1. Substitute the function f(x) = 4x^3 + ax^2 + bx + k into the integral:
∫ (from 0 to 1) (4x^3 + ax^2 + bx + k) dx = 32
2. Integrate each term separately and apply the limits:
∫ (4x^3 dx) + ∫ (ax^2 dx) + ∫ (bx dx) + ∫ (k dx) = 32
3. Evaluate each integral and substitute the limits:
[x^4 + (a/3)x^3 + (b/2)x^2 + kx] (from 0 to 1) = 32
4. Apply the limits:
[(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
Simplifying further, we get:
[1 + (a/3) + (b/2) + k] - [0 + 0 + 0 + 0] = 32
This gives us the equation:
1 + (a/3) + (b/2) + k = 32
Rearrange the equation to solve for k:
k = 32 - 1 - (a/3) - (b/2)
Substitute the values of a = 24 and b = 36:
k = 32 - 1 - (24/3) - (36/2)
k = 32 - 1 - 8 - 18
k = 5
Therefore, the value of k is 5.