Given:
![\begin{gathered} f(x)=\sqrt[3]{x} \\ \\ \text{ Tangent line t f\lparen x\rparen at x = 27} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ktck2q7mpa81r44zz3u2ezo6a27t1xpdog.png)
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):

• Now, let's find f'(27):
![\begin{gathered} f^(\prime)(27)=(1)/(3)(27)^{-(2)/(3)} \\ \\ f^(\prime)(27)=\frac{(1)/(3)}{\sqrt[3]{27}^2}=((1)/(3))/(3^2)=((1)/(3))/(9)=(1)/(3*9)=(1)/(27) \\ \\ f^{^(\prime)}(27)=(1)/(27) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7296z7yondgwx0yoj2x3wqk3izzazrbwr1.png)
Thus, we have:
Slope, m = 1/27
• The equation of through (27, 3) and has a slope of 1/27 will be:

Therefore, the equation of the tangent line is:

To find the approximation, we have:
![\begin{gathered} \sqrt[3]{27.2}=f(27.2) \\ \\ f(x)=(1)/(27)x+2 \\ Where\text{ x = 27.2} \\ Substitute\text{ 27.2 for x.} \\ \\ f(27.2)=(1)/(27)(27.2)+2 \\ \\ f(27.2)=(27.2)/(27)+2 \\ \\ f(27.2)=1.0074+2 \\ \\ f(27.2)=3.0074 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7wh5h92kqrzhef7c6fb9mcse0944wg5lts.png)
ANSWER:
![\begin{gathered} m=(1)/(27) \\ \\ b=2 \\ \\ \sqrt[3]{27.2}=3.0074 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hog09to2l2msauiju4ig6m85doxwqshtul.png)