Question

Sketch the graph of the function. Choose a scale that allows all relative extrema and points of inflection to be identified on the graph. \(y=x^{4}-4 x^{3}+16 x\)

Short answer

The function \(y=x^{4}-4 x^{3}+16 x\) has relative extrema and points of inflection at \(x=0\) and \(x=2\). At \(x=0\), \(y=16\) and at \(x=2\), \(y=8\).

Step-by-step solution

Compute the First Derivative

Find the first derivative of \(y=x^{4}-4x^{3}+16x\). Using the power rule (\(d/dx[x^n] = n*x^{n-1}\)), the first derivative of the function is \( y'= 4x^{3} - 12x^{2} + 16 \).

Calculate Critical Points

Set the first derivative equal to zero to find the critical points: \(4x^{3} - 12x^{2} + 16 = 0\). Solving this gives points \(x = 0, 2\). These are the points where the function has a potential maximum or minimum.

Find the Second Derivative

Calculate the second derivative by taking the derivative of the first derivative: \(y''=12x^{2}-24x\). Set the second derivative equal to zero to find potential points of inflection: \(12x^{2}-24x = 0\). Solving this yields \(x = 0, 2\), which are also the points of inflection where the curvature of the graph changes.

Summarize Key Points

For \(x=0\) and \(x=2\), we have relative extrema because the first derivative changes sign. For the same points, the second derivative also changes sign, making these points the points of inflection as well.

Graph The Function

Use all the identified information to sketch the graph. The y-intercept is at \(y=16\) when \(x=0\). The curve changes slope at points \(x=0\) and \(x=2\). At \(x=0\) and \(x=2\), the curve changes its concavity as well. Hence relative extrema and inflection points, all occur at the same points.