Suppose f"(x)=6x-2
A. If f'(1)=5, find f'(x).
I got f'(x)=3x^2-2x+4
B. Fine the average rate of change of f on [-1,4].
C. Fine the value of x guaranteed by the mean value theorem for r on [-1,4].
D. If f(-1)=-2, find f(x)
I am not sure how to do B, C, and D.
the average rate of change of f on [a,b] is just [f(b)-f(a)]/(b-a)
f(x) = x^3-x^2+4x+c
The c does not matter, so the average rate of change on [-1,4] is just
[f(4)-f(-1))/3 = ((64-16+16+c)-(-1-1-4+c))/3 = 70/3
You want a c where f'(c) = 70/3
3c^2-2c+4 = 70/3
c = (1+√59)/3 = 2.89
which is in [-1,4]
since you know f(-1), you have
f(-1) = -1-1-4+c = -2
c = 4
as always, double-check my math - this was done on the fly.
B. To find the average rate of change of f on the interval [-1,4], we can use the formula:
Average Rate of Change = (f(b) - f(a)) / (b - a)
where a = -1 and b = 4.
To calculate f(b) and f(a), we need to find the antiderivative of f"(x). Integrating f"(x) once gives:
f'(x) = ∫(6x - 2) dx
f'(x) = 3x^2 - 2x + C
Using the given condition f'(1) = 5, we can solve for C:
5 = 3(1)^2 - 2(1) + C
C = 5 - 3 + 2
C = 4
So, f'(x) = 3x^2 - 2x + 4.
Now, substitute the values into the average rate of change formula:
Average Rate of Change = (f(4) - f(-1)) / (4 - (-1))
To find the values of f(4) and f(-1), we need to integrate f'(x) again:
f(x) = ∫(3x^2 - 2x + 4) dx
f(x) = x^3 - x^2 + 4x + D
Using the given condition f(-1) = -2, we can solve for D:
-2 = (-1)^3 - (-1)^2 + 4(-1) + D
D = -2 + 1 - 4 - (-1)
D = -2 + 1 - 4 + 1
D = -4
So, f(x) = x^3 - x^2 + 4x - 4.
Now, substitute the values into the average rate of change formula:
Average Rate of Change = (f(4) - f(-1)) / (4 - (-1))
= (4^3 - 4^2 + 4(4) - 4) - ((-1)^3 - (-1)^2 + 4(-1) - 4)) / (4 - (-1))
= (64 - 16 + 16 - 4) - (1 - 1 - 4 - 4) / (4 + 1)
= (60 - 1) / 5
= 59 / 5
= 11.8
Therefore, the average rate of change of f on the interval [-1,4] is 11.8.
C. The Mean Value Theorem states that if a function f(x) is continuous on the interval [a,b] and differentiable on (a,b), then there exists a value c in the interval (a,b) such that:
f'(c) = (f(b) - f(a)) / (b - a)
In this case, a = -1 and b = 4. We already have the values for f(a) and f(b):
f(a) = f(-1) = -2
f(b) = f(4) = 60
To find the value of c guaranteed by the Mean Value Theorem, we need to solve the equation:
f'(c) = (f(b) - f(a)) / (b - a)
3c^2 - 2c + 4 = (60 - (-2)) / (4 - (-1))
3c^2 - 2c + 4 = 62 / 5
Multiplying both sides of the equation by 5 to remove the fraction gives:
15c^2 - 10c + 20 = 62
Rearranging the equation gives:
15c^2 - 10c - 42 = 0
Now we can solve this quadratic equation for c using the quadratic formula:
c = (-(-10) ± √((-10)^2 - 4(15)(-42))) / (2(15))
c = (10 ± √(100 + 2520)) / 30
c = (10 ± √2620) / 30
Simplifying the square root:
c = (10 ± √4 * 655) / 30
c = (10 ± 2√655) / 30
So, the value of x guaranteed by the Mean Value Theorem for f on the interval [-1,4] is approximately:
c ≈ (10 ± 2√655) / 30
D. The function f(x) = x^3 - x^2 + 4x - 4 was found in part B. To find f(x) when f(-1) = -2, substitute x = -1 into the equation:
f(x) = (-1)^3 - (-1)^2 + 4(-1) - 4
= -1 + 1 - 4 - 4
= -8
So, when f(-1) = -2, f(x) = -8.
B. To find the average rate of change of f on the interval [-1, 4], use the formula:
Average Rate of Change = (f(4) - f(-1)) / (4 - (-1))
In this case, we need to find the values of f(4) and f(-1) first.
C. The Mean Value Theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in (a, b) such that:
f'(c) = (f(b) - f(a)) / (b - a)
In this case, the interval is [-1, 4]. We need to find the values of f(-1) and f(4) first.
D. To find f(x) when f(-1) = -2, you need to integrate f''(x) twice.
First, integrate f''(x) with respect to x to find f'(x). Since f''(x) = 6x - 2, integrating once gives:
f'(x) = 3x^2 - 2x + C₁
Next, integrate f'(x) with respect to x to find f(x). Since f'(x) = 3x^2 - 2x + C₁, integrating again gives:
f(x) = x^3 - x^2 + C₁x + C₂
To find the specific function f(x) given f(-1) = -2, substitute x = -1 and f(-1) = -2 into the equation above and solve for the constants C₁ and C₂.