Problem 66 - College Algebra and Calculus: An Applied Approach

Question

Use a graphing utility to graph \(f\) and \(f^{\prime}\) over the given interval. Determine any points at which the graph of \(f\) has horizontal tangents. $$ f(x)=x^{3}-1.4 x^{2}-0.96 x+1.44 \quad[-2,2] $$

Step-by-Step Solution

Answer

The solutions for \(x\) which are in the interval \([-2,2]\) are the \(x\)-coordinates of the points where \(f(x)\) has a horizontal tangent.

Step 1: Graph the function \(f(x)\)

First, a graphing utility can be used to plot the given function \(f(x) = x^{3} - 1.4x^{2} - 0.96x + 1.44\) over the specified interval of \([-2, 2]\). This will provide a visual representation of the function. Note the overall shape and any major points of interest.

Step 2: Find the derivative \(f^{\prime}(x)\)

The derivative of the function, \(f^{\prime}(x)\), represents the slope of the tangent line at any point along the original function \(f(x)\). Use the power rule to take the derivative: for each term, multiply the coefficient by the current power of \(x\), and then decrease that power by 1. Doing so in this case gives: \(f^{\prime}(x) = 3x^2 - 2.8x - 0.96\).

Step 3: Solve \(f^{\prime}(x) = 0\) for \(x\)

The derivative is equal to zero at points of horizontal tangency on the graph of \(f(x)\). Therefore, the values of \(x\) that solve the equation \(f^{\prime}(x) = 0\) represent the \(x \)-coordinates of this horizontal tangency. The roots of this equation \(3x^2 - 2.8x - 0.96 = 0\) can be found by applying the quadratic formula, \(x = [-b \pm \sqrt{b^2 - 4ac}]/(2a)\), where \(a = 3\), \(b = -2.8\) and \(c = -0.96\).

Step 4: Check the solutions in the given interval

After obtaining the solutions for \(x\) from the quadratic formula, check to see if these \(x\) values are within the given interval \([-2, 2]\). If they are, they represent valid points of horizontal tangency.