Problem 43 - 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. $$ f(x)=\sqrt{x}+1 ;(4,3) $$

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = \frac{1}{4}x + 2\). The slope of this line and tangent point (4,3) visually matches the function graph when plotted on a graphing utility.

Step 1: Finding the Derivative

The slope of the tangent line at a particular point is equal to the derivative of the function at that point. So, first find the derivative of \(f(x)\) using the limit definition of derivative:\[f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}\] Substitute \(f(x)=\sqrt{x}+1\), \(x=4\), and \(h\) tending to 0.

Step 2: Simplification of the Derivative

With \(f(x+h)\) in the numerator, substitute \(f(x)\) and simplify the expression to isolate the limit. The progress of simplification should be as follows:\[ = \lim_{h\to 0}\frac{\sqrt{4+h}+1 - (\sqrt{4}+1)}{h}\]\[ = \lim_{h\to 0}\frac{\sqrt{4+h}-2}{h}\] After simplification, calculate the limit as \(h\) approaches 0. This will give the slope of the tangent line.

Step 3: Finding Point Slope Form

Now having the slope of the tangent line from the previous step, plug it into the equation of a line, \(y-y1=m(x-x1)\), where \((x1,y1)\) are coordinates for a point on the line and \(m\) is the slope. In this case, \((x1,y1) = (4,3)\), and \(m\) is the limit calculated.

Step 4: Verify the Results Graphically

Lastly, using a graphing utility, plot the equation of the tangent line and the function to visually verify the results. The tangent line should exactly touch the function graph at the given point (4,3).