Problem 43 - College Algebra and Calculus: An Applied Approach

Question

Find an equation of the tangent line to the graph of \(f\) at the point \((2, f(2)) .\) Use a graphing utility to check your result by graphing the original function and the tangent line in the same viewing window. $$ f(x)=\sqrt{4 x^{2}-7} $$

Step-by-Step Solution

Answer

The equation of the tangent line to the graph of the function \(f(x) = \sqrt{4x^2 - 7}\) at the point \(2, f(2)\) is given by \(y = \frac{8}{3}x - \frac{7}{3}\).

Step 1: Transform the function

Change the form of function \(f(x)=\sqrt{4x^2-7}\) into a power for easier differentiation. This results in \(f(x) = (4x^2 -7)^{0.5}\).

Step 2: Calculate derivative

Next, find the derivative of \(f(x)\). Using the chain rule, the derivative \(f'(x)\) is \( \frac{1}{2}(4x^2-7)^{-1/2} \cdot 8x = \frac{4x}{\sqrt{4x^2-7}}\).

Step 3: Find slope of tangent line

To get the slope of the tangent line at \(x=2\), plug this value into \(f'(x)\). Thus, the derivative at \(x = 2\) is \(f'(2) = \frac{4 \cdot 2}{\sqrt{4 \cdot 2^2-7}} =\frac{8}{\sqrt{9}} = \frac{8}{3}\). So, the slope of the tangent at the point is \(8/3\).

Step 4: Use point-slope form to get the equation

Using the point-slope form of a linear equation (y - y1 = m(x - x1)), where m is the slope of the tangent line and (x1, y1) is the point of tangency (2, f(2)), we get the tangent line equation as \(y - f(2) = \frac{8}{3} (x - 2)\). Substituting for \(f(2)\), which equals \(\sqrt{4 \cdot 2^2 - 7} = \sqrt{9} = 3\), we get the equation of the tangent line \(y - 3 = \frac{8}{3}(x - 2)\). By algebraically rearranging, this leads to the equation of the tangent line: \(y = \frac{8}{3}x - \frac{7}{3}\).

Step 5: Check with a graphing utility

To confirm correctness, graph \(f(x) = (4x^2 -7)^{0.5}\) and the found tangent line \(y = \frac{8}{3}x - \frac{7}{3}\) in the same window of a graphing utility. The tangent line should just touch the graph of the function at the point (2, f(2)).