Question

Use linear approximation, i.e. the tangent line, to approximate sqrt(36.2) as follows Let f(x) = sqrt(x) Find the equation of the tangent line to f(x) at x = 36

Answer 1

Answer: 6.01666667

Explanation:

The slope of the line tangent to f(x)=√x is the derivative of the function which is:

$f^{\prime}(x)=\frac{1}{2\sqrt{x}}$

This means that at x=36, the slope is:

$\begin{gathered} f^{\prime}(x)=\frac{1}{2\sqrt{x}} \\ f^{\prime}(36)=\frac{1}{2\sqrt{36}}=\frac{1}{12} \end{gathered}$

The function value for x=36 is:

$\begin{gathered} f(x)=\sqrt{x} \\ f(36)=\sqrt{36}=6 \end{gathered}$

This means that the point on the curve we are being asked to find the equation of the tangent line would be (36, 6). Using the point-slope form:

$\begin{gathered} y-6=\frac{1}{12}(x-36) \\ y-6=\frac{1}{12}x-3 \\ y=\frac{1}{12}x-3+6 \\ y=\frac{1}{12}x+3 \\ L(x)=\frac{1}{12}x+3 \end{gathered}$

With this, the linear approximation of √36.2 would be:

$\begin{gathered} L(x)=\frac{1}{12}x+3 \\ L(36.2)=\frac{1}{12}(36.2)+3 \\ L(36.2)=6.01666667 \end{gathered}$