Question

In Exercises 39 and \(40,\) use the derivative of the function \(y=f(x)\) to find the points at which \(\int\) has a (a) local maximum, (b) local minimum, or (c) point of inflection. $$y^{\prime}=(x-1)^{2}(x-2)(x-4)$$

Short answer

The points where \(y\) has local maximum, minimum or points of inflection are \(x = 1, 2, 4\). The local minimums are at \(x = 1\) and \(4\), the local maximum is at \(x = 2\), and the points of inflection are at \(x = 1.5, 3, 5\).

Step-by-step solution

Find the Derivative

The given derivative function is \(y' = (x-1)^2(x-2)(x-4)\). In order to find the points where the integral has local maximum, minimum or points of inflection, one needs to set the derivative equal to zero: \((x-1)^2(x-2)(x-4) = 0\).

Find the Points

By solving the above equation, four roots will be obtained corresponding to \(x = 1,2,4\). The power of 2 for the first term means \(x = 1\) is a repeated root.

Test each interval

Use the first and second derivative tests to find the local maxima, minima and inflection points. The first derivative gives the rate of change; when the sign changes from positive to negative, it indicates a local maximum point and vice versa for a local minimum point. The second derivative tells about concavity; a change from negative to positive suggests a point of inflection. For \(x > 4, y' > 0\), for \(2 < x < 4, y' < 0\), for \(1 < x < 2, y' > 0\), and for \(x < 1, y' > 0\). Thus, \(y\) decreases for \(1 < x < 2\) and \(2 < x < 4\) then increases for \(x < 1\) and \(x > 4\). We can thus infer that \(x = 1\) and \(4\) are points of local minimum, while \(x = 2\) is a point of local maximum.

Identify Points of Inflection

To find points of inflection, it is necessary to find the second derivative and set it to zero. Obtaining the second derivative of \(y\), setting it to equal to zero, and then solving for \(x\), yields \(x = 1.5, 3, 5\). Thus, the points of inflection are at \(x = 1.5, 3, 5\).