Question

question 20 · 1 point given the function of ( f(x)=7x^{1/2}-3 ) on the interval (9,36), find a point ( c ) between 9 and 36 so that it satisfies the mean - value theorem. provide your answer below: ( c=square )

Explanation:

Step1: Recall Mean - Value Theorem formula

The Mean - Value Theorem states that if \(y = f(x)\) is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), then \(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\), where \(a = 9\), \(b = 36\), and \(f(x)=7x^{\frac{1}{2}}-3\). First, find \(f(a)\) and \(f(b)\).
\[f(9)=7\times9^{\frac{1}{2}}-3=7\times3 - 3=21 - 3=18\]
\[f(36)=7\times36^{\frac{1}{2}}-3=7\times6 - 3=42 - 3=39\]
Then \(\frac{f(36)-f(9)}{36 - 9}=\frac{39 - 18}{27}=\frac{21}{27}=\frac{7}{9}\)

Step2: Find the derivative of \(f(x)\)

Differentiate \(f(x)=7x^{\frac{1}{2}}-3\) using the power - rule \((x^n)^\prime=nx^{n - 1}\). So \(f^{\prime}(x)=\frac{7}{2}x^{-\frac{1}{2}}\)

Step3: Set \(f^{\prime}(c)\) equal to \(\frac{f(36)-f(9)}{36 - 9}\) and solve for \(c\)

Set \(\frac{7}{2}c^{-\frac{1}{2}}=\frac{7}{9}\)
Cross - multiply to get \(7\times9 = 7\times2c^{\frac{1}{2}}\)
Divide both sides by 7: \(9 = 2c^{\frac{1}{2}}\)
Then \(c^{\frac{1}{2}}=\frac{9}{2}\)
Square both sides: \(c=\frac{81}{4}=20.25\)

Answer:

\(20.25\)