Find a point c satisfying the conclusion of the Mean Value Theorem for the following function and interval.
f(x)=x^−1 [1,9]
Well, finding a point that satisfies the conclusion of the Mean Value Theorem for the given function and interval is quite intriguing. Why don't we delve into the madness of mathematical merriment together?
Let's start by recalling the Mean Value Theorem. It states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in the open interval (a, b) such that the instantaneous rate of change (i.e., derivative) is equal to the average rate of change over the interval.
Now, for the function f(x) = x^(-1) on the interval [1,9], we need to check if it satisfies the conditions of the Mean Value Theorem. Since f(x) is continuous on the closed interval [1,9] and differentiable on the open interval (1,9), we can proceed with our quest for the elusive point c.
To find the point c, we need to determine the average rate of change of f(x) over the interval [1,9]. The average rate of change is simply the difference in the function values divided by the difference in the inputs. So let's compute:
Average rate of change = (f(9) - f(1)) / (9 - 1)
= (1/9 - 1/1) / 8
= (-8/9) / 8
= -1/9
Now, we need to find a point c in the open interval (1,9) such that the derivative of f(x) at that point is equal to the average rate of change. Let's find the derivative of f(x):
f'(x) = -1 * x^(-2)
= -1/x^2
To find c, we set f'(c) equal to the average rate of change:
-1/c^2 = -1/9
Cross-multiplying, we get:
9 = c^2
Taking the square root of both sides, we find:
c = ±3
So, we have two potential points: c = 3 and c = -3. However, since we need to find a point in the open interval (1,9), only c = 3 satisfies this condition.
Therefore, the point c that satisfies the conclusion of the Mean Value Theorem for the function f(x) = x^(-1) on the interval [1,9] is c = 3.
And just like that, we've landed in the magical realm of mathematical humor. I hope you enjoyed this mathematical journey, and may laughter always be your greatest derivative!
To find a point that satisfies the conclusion of the Mean Value Theorem for the function f(x) = x^(-1) over the interval [1,9], we first need to verify if the function fulfills the requirements of the Mean Value Theorem.
The Mean Value Theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
In this case, the function f(x) = x^(-1) is continuous and differentiable on the interval (1, 9), but it is not differentiable at x = 0 since the derivative does not exist there. Thus, we cannot find a point c satisfying the conclusion of the Mean Value Theorem for the given function and interval.
Hope this helps! Let me know if you have any further questions.
To find a point c satisfying the conclusion of the Mean Value Theorem for the function f(x) = x^(-1) on the interval [1,9], we need to check if the function satisfies the conditions of the Mean Value Theorem.
According to the Mean Value Theorem, for a function that is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), there exists at least one point c in (a, b) such that the slope of the tangent line at c is equal to the average rate of change of the function over the interval [a, b].
First, let's check the continuity of f(x) = x^(-1) on the interval [1,9]. The function is continuous everywhere except x = 0, but since the interval does not include 0, f(x) is continuous on [1,9].
Next, let's check the differentiability of f(x) = x^(-1) on the open interval (1,9). To do this, we need to find the derivative of f(x) and check if it is defined and continuous on (1,9).
The derivative of f(x) = x^(-1) can be found by applying the power rule, which states that d/dx x^n = n*x^(n-1). Applying this rule, we get:
f'(x) = d/dx (x^(-1)) = -1*x^(-2) = -1/x^2
Since 1/x^2 is defined and continuous for all x ≠ 0, we can conclude that f(x) = x^(-1) is differentiable on the open interval (1,9).
Now that we have established the continuity and differentiability of f(x) = x^(-1) on the appropriate intervals, we can use the Mean Value Theorem to find a point c where the slope of the tangent line is equal to the average rate of change.
The average rate of change of f(x) over the interval [1,9] can be found by calculating:
Average rate of change = (f(9) - f(1))/(9 - 1) = (1/9 - 1)/(8) = -8/9
To find the point c, we need to solve:
f'(c) = -8/9
Substituting f'(x) = -1/x^2, we get:
-1/c^2 = -8/9
Cross-multiplying, we have:
9 = -8c^2
Dividing both sides by -8, we get:
c^2 = -9/8
This equation has no real solutions since the square of a real number cannot be negative. Therefore, there is no point c on the interval (1,9) that satisfies the conclusion of the Mean Value Theorem for the function f(x) = x^(-1) on the interval [1,9].
f(1)=1/1=1
f(9)=1/9
The mean slope is ((1/9)-1)/(9-1)= -1/9
Now calculate f'(x)=-1/x²
So we need to solve for
f'(x)=-1/9
which gives x=3 or -3.
Since only x=3 is on the interval [1,9], this is the only solution, i.e. c=3.