Problem 42 - 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. $$ f(x)=2 x^{2}-1 ;(0,-1) $$

Step-by-Step Solution

Answer

Using the limit definition of the derivative, the slope of the tangent line to \(f(x) = 2x^2 - 1\) at point (0, -1) is 0. Thus, the equation of the tangent line at that point is y = -1.

Step 1: Compute derivative using limit definition

The limit definition of the derivative is given by: \[ f'(a)=\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}\] So, for \(f(x)=2x^2-1\) at point (0,-1): Using f(x) at x = 0+h and x = 0, \(f(0+h)=2(0+h)^2-1 =2h^2-1\) and \(f(0)=2(0)^2-1= -1\). So, \[f'(0)=\lim_{h \to 0} \frac{2h^2-1-(-1)}{h} =\lim_{h \to 0} 2h =0\]

Step 2: Find equation of the tangent line

The equation of a line is given by the point-slope form: \[y-y_{1}=m(x-x_{1})\] Using the calculated derivative f'(0)=0 as the slope m at the point (0,-1), the equation of the tangent line is: \[y-(-1)=0 \cdot (x-0)\] or simply \[y= -1\]

Step 3: Verification using a graphing tool

To verify the solution, graph the function and the equation of the tangent line in a graphing tool. The graph of y = \(2x^2 -1\) should have a tangent line at the point (0, -1) with the equation y = -1.