Use a linear approximation to estimate the number (125.07)2/3
Let \(f(x)=x^{2/3}\). You have \(f(125)=(125)^{2/3}=(5^3)^{2/3}=5^2=25\).
Taking the derivative, you get
\(f'(x)=\dfrac2{3x^{1/3}}\implies f'(125)=\dfrac2{3(125)^{1/3}}=\dfrac2{3\times5}=\dfrac2{15}\)
The linear approximation to \(f(x)\) near \(x=125\) is given by
\(L(x)=f(125)+f'(125)(x-125)\approx f(x)\)
So the value of \((125.07)^{2/3}\) can be estimated as
\(L(125.07)=f(125)+f'(125)(125.07-125)\)
\(L(125.07)=25+\dfrac2{15}(0.07)=25.009333\ldots\)
The estimated value of the given number is 25.0093.
Given:
The given number is \((125.07)^{\frac{2}{3}}\).
To find:
The estimated value of the given number by using the linear approximation formula.
Explanation:
Linear approximation formula:
\(f(x)=f(x_0)+f'(x_0)(x-x_0)\)
Let
\(f(x)=x^{\frac{2}{3}}\)
For \(x_0=125\),
\(f(125)=(125)^{\frac{2}{3}}\)
\(f(125)=5^2\)
\(f(125)=25\)
Differentiate \(f(x)\) with respect to \(x\).
\(f'(x)=\dfrac{2}{3}x^{\frac{2}{3}-1}\)
\(f'(x)=\dfrac{2}{3}x^{-\frac{1}{3}}\)
\(f'(x)=\dfrac{2}{3x^{\frac{1}{3}}}\)
For \(x_0=125\),
\(f'(125)=\dfrac{2}{3(125)^{\frac{1}{3}}}\)
\(f'(125)=\dfrac{2}{3(5)}\)
\(f'(125)=\dfrac{2}{15}\)
Using the linear approximation formula, we get
\(f(125.07)=f(125)+f'(125)(125.07-125)\)
\((125.07)^{\frac{2}{3}}=25+\dfrac{2}{3}(0.07)\)
\((125.07)^{\frac{2}{3}}=25.0093333...\)
\((125.07)^{\frac{2}{3}}\approx 25.0093333...\)
Therefore, the estimated value of the given number is 25.0093.
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