Questions: Let g(x)=-x^3+48x (a) Determine whether g is even, odd, or neither. (b) There is a local minimum value of -128 at -4. Determine the local maximum value.

Solution Steps

To determine whether the function \( g(x) = -x^3 + 48x \) is even, odd, or neither, we need to check the symmetry properties of the function. A function is even if \( g(-x) = g(x) \) for all \( x \), and it is odd if \( g(-x) = -g(x) \) for all \( x \).

To find the local maximum value, we need to find the critical points by taking the derivative of \( g(x) \), setting it to zero, and solving for \( x \). Then, we can use the second derivative test to determine the nature of these critical points.

To determine if the function \( g(x) = -x^3 + 48x \) is even, odd, or neither, we evaluate \( g(-x) \) and compare it to \( g(x) \) and \(-g(x)\).

• Calculate \( g(-x) \): \[ g(-x) = -(-x)^3 + 48(-x) = x^3 - 48x \]

• Compare \( g(-x) \) with \( g(x) \) and \(-g(x)\): \[ g(-x) = x^3 - 48x \quad \text{and} \quad -g(x) = x^3 - 48x \]

Since \( g(-x) = -g(x) \), the function \( g(x) \) is odd.

To find the critical points, we take the derivative of \( g(x) \) and set it to zero:

• Derivative of \( g(x) \): \[ g'(x) = -3x^2 + 48 \]

• Set \( g'(x) = 0 \) and solve for \( x \): \[ -3x^2 + 48 = 0 \quad \Rightarrow \quad x^2 = 16 \quad \Rightarrow \quad x = \pm 4 \]

The critical points are \( x = -4 \) and \( x = 4 \).

To determine the nature of the critical points, we use the second derivative test:

• Second derivative of \( g(x) \): \[ g''(x) = -6x \]

• Evaluate \( g''(x) \) at the critical points:

  • At \( x = -4 \): \[ g''(-4) = -6(-4) = 24 \quad (\text{local minimum}) \]
  • At \( x = 4 \): \[ g''(4) = -6(4) = -24 \quad (\text{local maximum}) \]

Since \( g''(4) < 0 \), there is a local maximum at \( x = 4 \).

To find the local maximum value, evaluate \( g(x) \) at \( x = 4 \):

• Calculate \( g(4) \): \[ g(4) = -(4)^3 + 48(4) = -64 + 192 = 128 \]

Final Answer