Question

In Exercises \(1-6,\) use the First Derivative Test to determine the local extreme values of the function, and identify any absolute extrema. Support your answers graphically. $$y=-2 x^{3}+6 x^{2}-3$$

Short answer

The function \(y=-2x^{3}+6x^{2}-3\) has a local maximum at \(x=0\), a local minimum at \(x=2\), and an absolute minimum of \(y=-5\) at \(x=2\). The function has no absolute maximum.

Step-by-step solution

Calculate the Derivative

Differentiate the function \(y=-2x^{3}+6x^{2}-3\) to get the derivative: \(y'=-6x^{2}+12x\).

Find Critical Points

Set the derivative equal to zero and solve for x to get the critical points: \(-6x^{2}+12x=0\), which simplifies to \(x(x-2)=0\). This gives \(x=0\) and \(x=2\) as critical points.

Apply First Derivative Test

Use the first derivative test to classify these critical points as local maxima, minima, or neither. Using test points you will find that \(x=0\) is a local maximum and \(x=2\) is a local minimum.

Identify Absolute Extrema

The function is defined for all real numbers, so it has no endpoints. Since the function is a polynomial of degree 3, it approaches \(-\infty\) as \(x \rightarrow -\infty\), and approaches \(+\infty\) as \(x \rightarrow +\infty\). Therefore, the function has no absolute maximum, but the absolute minimum is \(y_{min}=y(2)=-5\).