Consider the function f(x) = 3 -2 x^2 on the interval [ -5 , 3 ].
(A) Find the average or mean slope of the function on this interval, i.e.
\displaystyle{\frac { f(3) - f(-5) }{ 3 - (-5) } = }
4
(B) By the Mean Value Theorem, we know there exists a c in the open interval (-5, 3) such that f'( c) is equal to this mean slope. For this problem, there is only one c that works. Find it.
c=__????
at 3
y = 3 - 2(9) = -15
at -5
y = 3 - 2(25) = -47
-15 + 47 = 32
32/(3+5) = 4 agree
dy/dx = -4 x
at our spot
-4x = 4
so x = -1 where the slope is 4
To find the value of c that satisfies the Mean Value Theorem for the function f(x) = 3 - 2x^2 on the interval [-5, 3], we need to find the derivative of the function first.
The derivative of f(x) with respect to x, denoted as f'(x), is obtained by differentiating each term of the function. In this case, the derivative is:
f'(x) = d/dx (3 - 2x^2)
To differentiate the function, we apply the power rule, which states that for any real number n, the derivative of x^n with respect to x is nx^(n-1).
Using the power rule, the derivative of -2x^2 with respect to x is -4x.
Therefore, the derivative of f(x) is:
f'(x) = -4x
Now that we have the derivative, we can use the Mean Value Theorem, which states that if a function is continuous over a closed interval [a, b] and differentiable over the open interval (a, b), then there exists at least one c in (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
In this case, we know the mean slope is 4, and we want to find the value of c that satisfies f'(c) = 4.
Setting f'(x) equal to 4, we have:
-4c = 4
Dividing both sides of the equation by -4, we get:
c = -1
Therefore, the value of c that satisfies the Mean Value Theorem for the function f(x) = 3 - 2x^2 on the interval [-5, 3] is c = -1.