the mean value theorem guarantees the existence of a special point on the graph of $y = \\sqrt{x}$ between…

the mean value theorem guarantees the existence of a special point on the graph of $y = \\sqrt{x}$ between $(0,0)$ and $(4,2)$. what are the coordinates of this point?\na $(2,1)$\nb $(1,1)$\nc $(2,\\sqrt{2})$\nd $(\\frac{1}{2},\\frac{1}{\\sqrt{2}})$\ne none of the above

the mean value theorem guarantees the existence of a special point on the graph of $y = \\sqrt{x}$ between $(0,0)$ and $(4,2)$. what are the coordinates of this point?\na $(2,1)$\nb $(1,1)$\nc $(2,\\sqrt{2})$\nd $(\\frac{1}{2},\\frac{1}{\\sqrt{2}})$\ne none of the above

Answer

Explanation:

Step1: Find the derivative of (y = \sqrt{x})

The function (y=\sqrt{x}=x^{\frac{1}{2}}). Using the power rule ((x^n)^\prime = nx^{n - 1}), we have (y^\prime=\frac{1}{2\sqrt{x}}).

Step2: Calculate the slope of the secant line

The slope of the secant line between the points ((0,0)) and ((4,2)) is (m=\frac{2 - 0}{4 - 0}=\frac{1}{2}).

Step3: Apply the Mean Value Theorem

By the Mean Value Theorem, (y^\prime(c)=\frac{1}{2\sqrt{c}}=\frac{1}{2}) (where (c) is the (x) - coordinate of the point). Solving the equation (\frac{1}{2\sqrt{c}}=\frac{1}{2}) for (c): Multiply both sides by (2\sqrt{c}) gives (1=\sqrt{c}), then (c = 1). When (x = 1), (y=\sqrt{1}=1).

Answer:

B. ((1,1))