Given:
[tex]\begin{gathered} f(x)=\sqrt[3]{x} \\ \\ \text{ Tangent line t f\lparen x\rparen at x = 27} \end{gathered}[/tex]Apply the slope-intercept form:
y = mx + b
Where m is the slope and b is the y-intercept.
• Let's find the derivative of f(x):
[tex]\begin{gathered} f^{-1}(x)=\frac{d}{dx}(x)^{\frac{1}{3}}=\frac{1}{3}(x^{\frac{1}{3}-1}) \\ \\ f^{-1}(x)=\frac{1}{3}x^{-\frac{2}{3}} \end{gathered}[/tex]• Now, let's find f'(27):
[tex]\begin{gathered} f^{\prime}(27)=\frac{1}{3}(27)^{-\frac{2}{3}} \\ \\ f^{\prime}(27)=\frac{\frac{1}{3}}{\sqrt[3]{27}^2}=\frac{\frac{1}{3}}{3^2}=\frac{\frac{1}{3}}{9}=\frac{1}{3*9}=\frac{1}{27} \\ \\ f^{^{\prime}}(27)=\frac{1}{27} \end{gathered}[/tex]Thus, we have:
Slope, m = 1/27
• The equation of through (27, 3) and has a slope of 1/27 will be:
[tex]\begin{gathered} y-3=\frac{1}{27}(x-27) \\ \\ y-3=\frac{1}{27}x-\frac{1}{27}*27 \\ \\ y-3=\frac{1}{27}x-1 \\ \\ y=\frac{1}{27}x-1+3 \\ \\ y=\frac{1}{27}x+2 \end{gathered}[/tex]Therefore, the equation of the tangent line is:
[tex]\begin{gathered} y=\frac{1}{27}x+2 \\ \\ \text{ Where:} \\ m\text{ = }\frac{1}{27} \\ \\ b=2 \end{gathered}[/tex]To find the approximation, we have:
[tex]\begin{gathered} \sqrt[3]{27.2}=f(27.2) \\ \\ f(x)=\frac{1}{27}x+2 \\ Where\text{ x = 27.2} \\ Substitute\text{ 27.2 for x.} \\ \\ f(27.2)=\frac{1}{27}(27.2)+2 \\ \\ f(27.2)=\frac{27.2}{27}+2 \\ \\ f(27.2)=1.0074+2 \\ \\ f(27.2)=3.0074 \end{gathered}[/tex]ANSWER:
[tex]\begin{gathered} m=\frac{1}{27} \\ \\ b=2 \\ \\ \sqrt[3]{27.2}=3.0074 \end{gathered}[/tex]