Problem 41 - 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. $$ f(x)=2\left(x^{2}-1\right)^{3} $$

Step-by-Step Solution

Answer

The equation of the tangent line to the graph of \(f(x)=2\left(x^{2}-1\right)^{3}\) at the point (2, f(2)) is \(y-16=144(x-2)\)

Step 1 - Find the derivative of the function

To begin, we first need to find the derivative of the function. This gives us the slope of the tangent line. The derivative of the given function \[f(x)=2\left(x^{2}-1\right)^{3}\] can be calculated using the Chain Rule for differentiation. The derivative is \[f'(x)= 2 \cdot 3\left(x^{2}-1\right)^{2} \cdot 2x = 12x\left(x^{2}-1\right)^{2}\]

Step 2 - Find the slope of the tangent line

We now need to find the slope of the tangent line to the graph at the point (2, f(2)). The slope of the tangent is just the value of the derivative at this point. Substituting x=2 into our derivative gives: \[f'(2) = 12 \cdot 2 \cdot ((2^2)-1)^2 = 144\] So the slope of the tangent line at (2, f(2)) is 144.

Step 3 - Find the y-coordinate of the point

Next, the y-coordinate at the point (2, f(2)) is calculated by substituting x=2 into our original function: \[f(2) = 2 \cdot ((2^2)-1)^3 = 16\] So the point is (2,16).

Step 4 - Get the equation of the tangent line

Finally, using the point-slope form of the line equation, we can find the equation of the tangent line. The point-slope form is given by: \[(y-y_1)=m(x-x_1)\] Substituting \(m=144\), \(x_1=2\) and \(y_1=16\) gives us the equation of the tangent line: \[y-16=144(x-2)\]