Problem 44 - College Algebra and Calculus: An Applied Approach

Question

Use the limit definition to find an equation of the tangent line to the graph of \(f\) at the given point. Then verify your results by using a graphing utility to graph the function and its tangent line at the point. $$ f(x)=\sqrt{x+2} ;(7,3) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the function \(f(x) = \sqrt{x+2}\) at the point (7,3) is \(y = \frac {x} {6} + 2\).

Step 1: Find the Derivative of the Function

The derivative of \(f(x) = \sqrt{x+2}\) can be found using the limit definition of a derivative, which is \(f'(x) = \lim_{h\to 0} \frac {f(x+h) - f(x)} {h}\). Substitution gives \(f'(x) = \lim_{h\to 0} \frac {\sqrt{x+h+2} - \sqrt{x+2}} {h}\). Simplify this by multiplying top and bottom by the conjugate to eliminate the square root in the numerator. This gives \(f'(x) = \lim_{h\to 0} \frac {h} {h(\sqrt{x+h+2} + \sqrt{x+2})} = \frac {1} {\sqrt{x+2} + \sqrt{x+2}} = \frac {1} {2\sqrt{x+2}}\).

Step 2: Evaluate the Derivative at the given point for the slope of the tangent

Evaluate this derivative at \(x=7\) to find the slope of the tangent line at the given point: \(f'(7) = \frac {1} {2\sqrt{7+2}} = \frac{1}{6}\). Thus, the slope of the tangent line at point (7, 3) is 1/6.

Step 3: Use the point-slope form to write the equation for the tangent line

Using the point-slope form of the line, \(y - y1 = m(x - x1)\), where \(m\) is the slope of the line, and \((x1, y1)\) is a point on the line, substitute \(m = \frac {1} {6}, x1 = 7\), and \(y1 = 3\) to get the equation: \(y - 3 = \frac {(x - 7)} {6}\). Simplify this to get \(y = \frac {x} {6} + 2\).

Step 4: Graph the function and the tangent line to verify the results

Plot both \(f(x) = \sqrt{x+2}\) and the tangent line \(y = \frac {x} {6} + 2\) on the same set of axes. The tangent line should touch the graph of \(f\) at exactly one point (7,3), showing that the results are correct.