Consider the function f(x)=4x^3–2x on the interval [–2,2]. Find the average or mean slope of the function on this interval. __14__
By the Mean Value Theorem, we know there exists at least one c in the open interval (–2,2) such that f(c) is equal to this mean slope.
For this problem, there are two values of c that work. The smaller one is_____ and the larger one is ______?
f(-2)=-32+4=-28
f(2)=32-4=28
f'(c)=(28-(-28))/(2-(-2))=14
f'(c)=12c^2-2=14
c^2=16/12=4/3
c=+-2/sqrt(3)
To find the mean slope of the function f(x) = 4x^3 – 2x on the interval [-2, 2], we need to calculate the average rate of change of the function over the interval.
The formula for the average rate of change on an interval [a, b] is:
Average Rate of Change = (f(b) - f(a)) / (b - a)
In this case, a = -2 and b = 2.
Average Rate of Change = (f(2) - f(-2)) / (2 - (-2))
Now, let's calculate f(2) and f(-2):
f(2) = 4(2^3) - 2(2) = 32 - 4 = 28
f(-2) = 4((-2)^3) - 2(-2) = -32 + 4 = -28
Substituting these values into the formula:
Average Rate of Change = (28 - (-28)) / (2 - (-2))
= 56 / 4
= 14
So, the average or mean slope of the function on the interval [-2, 2] is 14.
By the Mean Value Theorem, we know that there exists at least one c in the open interval (-2, 2) such that f(c) is equal to this mean slope.
To find the values of c, we need to solve the equation:
f'(c) = 14
To find the derivative of f(x), we differentiate the function:
f'(x) = 12x^2 - 2
Now, set f'(c) = 14:
12c^2 - 2 = 14
12c^2 = 16
c^2 = 16/12
c^2 = 4/3
Taking the square root of both sides:
c = ±√(4/3)
So, the smaller value for c is -√(4/3) and the larger value for c is √(4/3).
To find the average or mean slope of the function on the interval [–2,2], we need to find the slope between the two endpoints (-2 and 2).
The formula to calculate the slope between two points (x1, y1) and (x2, y2) is:
slope = (y2 - y1) / (x2 - x1)
In this case, the points are (-2, f(-2)) and (2, f(2)).
Let's calculate the slope:
Slope = (f(2) - f(-2)) / (2 - (-2))
To find f(2), substitute x = 2 into the function:
f(2) = 4(2)^3 - 2(2)
f(2) = 32 - 4
f(2) = 28
Similarly, to find f(-2), substitute x = -2 into the function:
f(-2) = 4(-2)^3 - 2(-2)
f(-2) = -32 + 4
f(-2) = -28
Now we can calculate the slope:
Slope = (28 - (-28)) / (2 - (-2))
Slope = (28 + 28) / 4
Slope = 56 / 4
Slope = 14
So, the average or mean slope of the function on the interval [-2, 2] is 14.
According to the Mean Value Theorem, since the average slope of the function on the interval exists, there must exist at least one value c in the open interval (-2, 2) such that f(c) is equal to this mean slope (14).
However, you asked for the two values of c. To find these values, we need to find when the derivative of the function is equal to 14.
The derivative of the function f(x) = 4x^3 - 2x can be found by differentiating each term separately. The power rule for derivatives states that if you have a term of the form a*x^n, the derivative is a*n*x^(n-1).
Differentiating 4x^3:
f'(x) = 4*3*x^(3-1)
f'(x) = 12x^2
Differentiating -2x:
f'(x) = -2*1
f'(x) = -2
Now we need to find the values of x where f'(x) = 14:
12x^2 - 2 = 14
Rearranging the equation:
12x^2 = 16
Dividing both sides by 12:
x^2 = 16/12
x^2 = 4/3
Taking the square root of both sides:
x = ±√(4/3)
Therefore, the two values of c that correspond to a mean slope of 14 are c = √(4/3) and c = -√(4/3).