17) Consider the function
f(x)=3x3−3x2+4x+2
Find the average slope of this function on the interval (−3,5).
By the Mean Value Theorem, we know there exists a c in the open interval (−3,5) such that f′(c) is equal to this mean slope. Find the two values of c in the interval which work, enter the smaller root first: __ ≤ __
To find the average slope of a function on an interval, we need to calculate the difference in the function's values at the endpoints of the interval and divide it by the difference in the x-coordinates.
1. Start by finding the values of the function at the endpoints of the interval:
f(-3) = 3(-3)^3 - 3(-3)^2 + 4(-3) + 2
= 3(-27) - 3(9) - 12 + 2
= -81 - 27 - 12 + 2
= -118
f(5) = 3(5)^3 - 3(5)^2 + 4(5) + 2
= 3(125) - 3(25) + 20 + 2
= 375 - 75 + 20 + 2
= 322
2. Calculate the difference in the function's values at the endpoints:
f(5) - f(-3) = 322 - (-118)
= 322 + 118
= 440
3. Calculate the difference in the x-coordinates:
5 - (-3) = 8
4. Divide the difference in function values by the difference in x-coordinates to get the average slope:
Average slope = (f(5) - f(-3)) / (5 - (-3))
= 440 / 8
= 55
Now, according to the Mean Value Theorem, there exists a value c in the open interval (-3, 5) such that f'(c) equals the average slope of 55. To find the specific values of c, we need to find the derivative of the function.
The derivative of f(x) = 3x^3 - 3x^2 + 4x + 2 is given by:
f'(x) = 9x^2 - 6x + 4
Now, we need to solve the equation f'(c) = 55 to find the values of c.
9c^2 - 6c + 4 = 55
Rearrange the equation:
9c^2 - 6c + 4 - 55 = 0
9c^2 - 6c - 51 = 0
Now, we can solve this quadratic equation to find the values of c.
Using any quadratic equation solving method (factoring, quadratic formula, completing the square, etc.), we find:
c = (-(-6) ± sqrt((-6)^2 - 4(9)(-51))) / (2(9))
Simplifying,
c = (6 ± sqrt(36 + 1836)) / 18
c = (6 ± sqrt(1872)) / 18
c = (6 ± 8√3) / 18
Simplifying further,
c = (1 ± 4√3) / 3
Therefore, the two values of c in the interval (-3, 5) that satisfy f'(c) = 55 are:
c1 = (1 - 4√3) / 3
c2 = (1 + 4√3) / 3
Enter the smaller root first:
c1 ≤ c2
(1 - 4√3) / 3 ≤ (1 + 4√3) / 3
To find the average slope of the function f(x) on the interval (-3, 5), we need to find the slope of the secant line between the two endpoint values of the interval.
The average slope formula is given by:
average slope = (f(b) - f(a)) / (b - a)
where a and b are the endpoint values of the interval.
In this case, a = -3 and b = 5. Let's calculate the average slope:
average slope = (f(5) - f(-3)) / (5 - (-3))
To calculate f(5), substitute x = 5 into the function:
f(5) = 3(5)^3 - 3(5)^2 + 4(5) + 2
Simplifying further:
f(5) = 375 - 75 + 20 + 2
f(5) = 322
To calculate f(-3), substitute x = -3 into the function:
f(-3) = 3(-3)^3 - 3(-3)^2 + 4(-3) + 2
Simplifying further:
f(-3) = -81 - 27 - 12 + 2
f(-3) = -118
Now we can substitute these values into the average slope formula:
average slope = (322 - (-118)) / (5 - (-3))
average slope = 440 / 8
average slope = 55
So, the average slope of the function on the interval (-3, 5) is 55.
Next, we can use the Mean Value Theorem to find the values of c in the open interval (-3, 5) such that f'(c) is equal to this mean slope.
The Mean Value Theorem states that if a function f(x) is continuous on the closed interval [a, b], differentiable on the open interval (a, b), then there exists at least one number c in (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
In this case, a = -3, b = 5, and the mean slope is 55.
To find the values of c, we need to find the derivative of f(x) and set it equal to the mean slope:
f'(x) = 9x^2 - 6x + 4
Setting f'(x) = 55, we get:
9x^2 - 6x + 4 = 55
Rearranging the equation:
9x^2 - 6x + 4 - 55 = 0
9x^2 - 6x - 51 = 0
We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 9, b = -6, and c = -51. Substituting these values into the quadratic formula:
x = (-(-6) ± √((-6)^2 - 4(9)(-51)))/(2(9))
Simplifying further:
x = (6 ± √(36 + 1836))/(18)
x = (6 ± √(1872))/(18)
x = (6 ± √(16 * 117))/(18)
x = (6 ± 4√(117))/(18)
Dividing numerator and denominator by 2:
x = (3 ± 2√(117))/(9)
Now we have two possible values for c:
c1 = (3 + 2√(117))/(9)
c2 = (3 - 2√(117))/(9)
Therefore, the two values of c in the interval (-3, 5) that work are:
c1 = (3 + 2√(117))/(9)
c2 = (3 - 2√(117))/(9)
So, the smaller root is c2 and the larger root is c1.