Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose \(h^{\prime}(x)=x^{2}(x-1)^{2}(x-2)\) and \(h(0)=0\). Sketch a graph of \(y=h(x)\).

Step-by-Step Solution

Answer

Graph passes through (0,0), inflects at x=1, and a minimum at x=2.

Step 1: Identify Critical Points

Find the critical points by setting the derivative to zero: \(h^{\prime}(x) = x^{2}(x-1)^{2}(x-2) = 0\). This implies that the critical points are \(x = 0\), \(x = 1\), and \(x = 2\).

Step 2: Analyze Critical Points

Determine the nature of each critical point by analyzing the factors of \(h^{\prime}(x)\). At \(x=0\), the factor \(x^{2}\) implies a local minimum or maximum. At \(x=1\), \((x-1)^{2}\) implies a point of inflection because the power is even. Finally, \(x=2\) has a single power, indicating a change in the direction of the slope.

Step 3: Determine Behavior at Each Critical Point

Test values around each critical point to determine the change in the function:- For \(x=0\), the function value changes from negative (to the left of 0) to zero at 0.- For \(x=1\), the derivative does not change sign, indicating a point of inflection.- At \(x=2\), the derivative changes from negative to positive, indicating a minimum.

Step 4: Use Given Condition

Utilize the constraint \(h(0) = 0\) to affirm the starting point of the function. This signifies that the graph passes through the origin.

Step 5: Sketch the Graph

Combine all the previous information:- The graph passes through the origin (0,0).- There is a local maximum or minimum at \(x=0\).- There is a point of inflection at \(x=1\).- There is a local minimum at \(x=2\).- Finally, analyze additional points if needed to ensure the correct curve from \(x=0\) to beyond \(x=2\). Draw the graph to reflect these findings.

Key Concepts

Critical Points

Critical points are where the derivative of a function equals zero or is undefined. These points are crucial because they can indicate places where the function's graph might change direction. For the given function, its derivative is: \[ h^{"}(x) = x^2(x-1)^2(x-2) \] To find the critical points, we set this equation to zero:

• \(x^2 = 0\) gives us \(x = 0\).
• \((x-1)^2= 0\) results in \(x = 1\).
• \(x-2 = 0\) yields \(x = 2\).

Identifying these points helps us determine where the slope of the function might be zero, indicating a potential change in the graph's shape.

Derivative Analysis

Analyzing the derivative f(x) gives us insight into the function's behavior. The derivative \(h'(x)\) provides information about the rate of change of function \(h(x)\). By looking at the factors of \(h'(x)\), we can tell whether a critical point is a maximum, minimum, or inflection point.

• The factor \(x^2\) at \(x=0\) suggests either a maximum or minimum because it has an even exponent.
• The factor \((x-1)^2\) at \(x=1\) indicates an inflection point, as it doesn't change sign over the point.
• The factor \(x-2\) results in a sign change at \(x=2\), implying a change in the slope direction.

By considering these factors, we can accurately anticipate the graph's behavior at these points.

Inflection Points

An inflection point occurs where a curve changes concavity, i.e., from concave up to concave down or vice versa. Unlike maxima and minima, where the derivative is zero, inflection points are marked by a zero second derivative but no change in sign.In our example, inspecting the factor \((x-1)^2\) shows that the derivative does not change sign around \(x=1\), so it's an inflection point. Here, the curve shifts from curving upwards to curving downwards or the other way round.

Local Minimum

A local minimum is a point where the function value is lower than any nearby points. To find a local minimum, look for places where the derivative changes from negative to positive. This indicates that before the point the slope was negative (going down) and after it's positive (going up). In the given function, at \(x=2\), the factor \(x-2\) signals a local minimum because the derivative switches from negative to positive as we pass through this point. Therefore the graph dips at \(x = 2\), creating a local trough.

Local Maximum

A local maximum is identified where the function reaches a peak relative to the nearby values. The derivative changes from positive to negative at a local maximum, showing that the slope rises, peaks, then falls.However, in this problem, \(x=0\) doesn't exhibit a clear local maximum due to the test condition \(h(0) = 0\). Instead, it simply indicates a transition. The function briefly changes direction as it passes through the point at the origin (0,0), but does not peak or dip further around this area.