Question

In Exercises, find all relative extrema of the function. Use the Second- Derivative Test when applicable. $$ f(x)=\frac{1}{2} x^{4}-\frac{1}{3} x^{3}-\frac{1}{2} x^{2} $$

Short answer

The function \(f(x)=\frac{1}{2} x^{4}-\frac{1}{3} x^{3}-\frac{1}{2} x^{2}\) has a relative maximum at \(x=0\) and relative minima at \(x=-1/2\) and \(x=1\).

Step-by-step solution

Compute the first derivative

Find the derivative of the given function \(f(x)=\frac{1}{2} x^{4}-\frac{1}{3} x^{3}-\frac{1}{2} x^{2}\). After applying the power rule for derivatives, which states that the derivative of \(x^n\), is \(n \cdot x^{n-1}\), the derivative \(f'(x)\) is \(2x^{3}-x^{2}-x\).

Find the critical points

The critical points of the function are found by setting the derivative equal to zero and solving for \(x\). So you get the equation \(2x^{3}-x^{2}-x=0\). This can be simplified by factoring out \(x\), which yields \(x(2x^{2}-x-1)=0\). Solving this equation yields \(x = 0\), \(x = -1/2\), and \(x = 1\).

Apply the Second Derivative Test

Find the second derivative of the function by differentiating the first derivative. The second derivative \(f''(x)\) is \(6x^2-2x-1\). Evaluate this at the critical points found in Step 2. For \(x=0\), \(f''(x)<0\), indicating a relative maximum. For \(x= -1/2\), \(f''(x)>0\), indicating a relative minimum. For \(x=1\), \(f''(x)>0\), indicating a relative minimum again.

Summary of the relative extrema

The function \(f(x)=\frac{1}{2} x^{4}-\frac{1}{3} x^{3}-\frac{1}{2} x^{2}\) has a relative maximum at \(x=0\) and relative minima at \(x=-1/2\) and \(x=1\)

Key concepts

Second-Derivative Test

One of the most reliable methods to determine whether a critical point is a relative maximum, minimum, or a saddle point is the Second-Derivative Test. This test involves taking the second derivative of the function, denoted as \(f''(x)\).

After finding the critical points by setting the first derivative \(f'(x)\) to zero, you check the concavity of the function at these points by plugging them into the second derivative. If \(f''(x) > 0\) at a critical point, the function is concave up, and the point is a relative minimum. Conversely, if \(f''(x) < 0\), the function is concave down, and the point is a relative maximum. If \(f''(x) = 0\), the test is inconclusive and you may need to use other methods to determine the nature of the critical point.

Critical Points

Critical points are where the function's behavior changes, meaning the derivative at those points is zero or undefined, signifying potential relative extrema (maxima or minima) or inflection points.

To find critical points, as we did in the exercise, you set the first derivative equal to zero and solve for \(x\). This will provide you with the potential locations of relative extrema. However, not all critical points yield relative extrema, which is why additional tests like the Second-Derivative Test are essential. In our case, solving \(2x^{3}-x^{2}-x=0\) led to \(x=0\), \(x=-1/2\), and \(x=1\) as the critical points where the function’s behavior could potentially change.

Power Rule for Derivatives

One of the most fundamental rules for finding derivatives is the Power Rule, which states that the derivative of \(x^n\) with respect to \(x\) is \(n \times x^{n-1}\). This rule makes it simpler to find the derivatives of polynomial functions.

When we applied this rule to our function \(f(x)=\frac{1}{2} x^{4}-\frac{1}{3} x^{3}-\frac{1}{2} x^{2}\), we differentiated each term independently to get the first derivative as \(f'(x)=2x^{3}-x^{2}-x\). Applying the Power Rule correctly is crucial because the correct form of the first derivative is what we use to determine the critical points and to perform the Second-Derivative Test.