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Sagot :
ANSWERS
• Linear approximation: ,y = (1/10)x + (22/5)
• (24.9)^1/2 ≈ 4.99
,• (25.1)^1/2 ≈ 5.01
EXPLANATION
To find the local linear approximation of the function, we have to find the equation of the tangent line of the function at x = 6.
The equation of a line with slope m and y-intercept b is,
[tex]y=mx+b[/tex]The slope of a function at a particular point is the derivative of the function evaluated in that point.
Let's find the derivative of f(x) using the exponent rule,
[tex]f^{\prime}(x)=\frac{1}{2}\cdot(19+x)^{(1/2)-1}=\frac{1}{2(19+x)^{1/2}}[/tex]Now, evaluate at x = 6,
[tex]f^{\prime}(6)=\frac{1}{2(19+6)^{1/2}}=\frac{1}{2(25)^{1/2}}=\frac{1}{2\sqrt[]{25}}=\frac{1}{2\cdot5}=\frac{1}{10}[/tex]For now, the equation of the line is,
[tex]y=\frac{1}{10}x+b[/tex]The function and the tangent line have the same value at the tangent point,
[tex]f(6)=(19+6)^{1/2}=(25)^{1/2}=\sqrt[]{25}=5[/tex]This means that the tangent point is (6, 5). We use this point to find the y-intercept of the tangent line,
[tex]5=\frac{1}{10}\cdot6+b[/tex]Solving for b,
[tex]b=5-\frac{6}{10}=\frac{22}{5}[/tex]The linear approximation of f(x) is,
[tex]y=\frac{1}{10}x+\frac{22}{5}[/tex]Now we want to use this to approximate (24.9)^1/2 and (25.1)^1/2. We have to use the values of x that, replacing in the function, would give the bases of these exponents:
[tex]\begin{gathered} 24.9=19+x\Rightarrow x=24.9-19=5.9 \\ \\ 25.1=19+x\Rightarrow x=25.1-19=6.1 \end{gathered}[/tex]To find each approximation, we have to replace x with each of these values in the linear approximation of f(x). The approximate values are:
[tex]\begin{gathered} (24.9)^{1/2}\approx\frac{1}{10}\cdot5.9+\frac{22}{5}=4.99 \\ \\ (25.1)^{1/2}\approx\frac{1}{10}\cdot6.1+\frac{22}{5}=5.01 \end{gathered}[/tex]
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