Consider the function f ( x ) = 3 − 5 x^2 on the interval [ − 2 , 8 ] . Find the average or mean slope of the function on this interval, i.e.
f(8)-f(-2)/8-(-2)=
By the Mean Value Theorem, we know there exists a c in the open interval ( − 2 , 8 ) such that f ' ( c ) is equal to this mean slope. For this problem, there is only one c that works. Find it.
Answer with explanation please and thanks!
really? They gave you the actual formula.
Just evaluate m = (f(8)-f(-2))/(8-(-2)) = ___
now, since f'(x) = -10x, you want to find c such that
-10c = m
c = -m/10
To find the average or mean slope of the function f(x) = 3 - 5x^2 on the interval [-2, 8], we need to calculate the difference in the function values divided by the difference in the x-values.
The difference in the function values is f(8) - f(-2):
f(8) = 3 - 5(8)^2 = 3 - 5(64) = 3 - 320 = -317
f(-2) = 3 - 5(-2)^2 = 3 - 5(4) = 3 - 20 = -17
So, f(8) - f(-2) = -317 - (-17) = -317 + 17 = -300
The difference in the x-values is 8 - (-2) = 8 + 2 = 10
Therefore, the average or mean slope of the function on the interval [-2, 8] is:
(-300) / (10) = -30
According to the Mean Value Theorem, there exists a point c in the open interval (-2, 8) where the derivative of the function, f'(c), is equal to this mean slope (-30).
To find this point c, we need to find the derivative of f(x) first:
f'(x) = d/dx (3 - 5x^2)
= 0 - 5(2x)
= -10x
Now we set f'(c) = -30 and solve for c:
-10c = -30
c = -30 / -10
c = 3
Therefore, the point c that satisfies the Mean Value Theorem is c = 3.
To find the average or mean slope of the function on the interval [−2, 8], we need to calculate the difference in y-values divided by the difference in x-values.
The formula for average slope is given by:
Average slope = (f(8) - f(-2)) / (8 - (-2))
Plugging in the values into the equation, we have:
Average slope = (f(8) - f(-2)) / (8 - (-2))
= (3 - 5(8)^2) - (3 - 5(-2)^2) / (8 - (-2))
= (3 - 320) - (3 - 20) / (8 + 2)
= (-317) - (-17) / (10)
= -300 / 10
= -30
So the average slope of the function on the interval [−2, 8] is -30.
Now, according to the Mean Value Theorem, there exists a c in the open interval (-2, 8) such that f'(c) (the derivative of f(x)) is equal to this mean slope.
To find this particular value of c, we need to find the derivative of the function f(x).
f(x) = 3 - 5x^2
To differentiate f(x) with respect to x, we use the power rule of differentiation:
f'(x) = d/dx (3) - d/dx (5x^2)
= 0 - 10x
= -10x
Now we set f'(c) equal to the average slope we found:
-10c = -30
Solving for c, we divide both sides of the equation by -10:
c = -30 / -10
= 3
Therefore, the value of c that satisfies f'(c) equals to the average slope of the function on the interval [−2, 8] is c = 3.