Question

For each of the following functions of one variable, determine all local maxima and minima and indicate points of inflection (where \(f^{\prime \prime}=0\) ): a. \(f(x)=4 x^{3}-12 x\) b. \(f(x)=4 x-x^{2}\) c. \(f(x)=x^{3}\)

Short answer

Question: Identify the local maxima, local minima, and points of inflection for the following functions: (a) \(f(x) = 4x^3 - 12x\), (b) \(f(x) = 4x - x^2\), and (c) \(f(x) = x^3\). Answer: (a) f(x) has a local maximum at x = -1, a local minimum at x = 1, and a point of inflection at x = 0. (b) f(x) has a local maximum at x = 2 and no point of inflection. (c) f(x) has no local maximum or minimum, but has a point of inflection at x = 0.

Step-by-step solution

Find the first derivative f'(x)

Compute the first derivative of the function using the power rule, \(f'(x) = 12x^2 - 12\).

Apply the first derivative test

Find critical points by setting the first derivative equal to zero and solving for x: \(12x^2 - 12 = 0\). Factor out 12: \(12(x^2 - 1) = 0\). This gives us the critical points: \(x = -1\) and \(x = 1\). Use the first derivative test to analyze these critical points: - For \(x < -1\), \(f'(-2) = 12\), so the function is increasing. - For \(-1 < x < 1\), \(f'(0) = -12\), so the function is decreasing. - For \(x > 1\), \(f'(2) = 12\), so the function is increasing. Therefore, we have a local maximum at \(x = -1\) and a local minimum at \(x = 1\).

Find the second derivative f''(x)

Compute the second derivative by differentiating f'(x), \(f''(x) = 24x\).

Determine the points of inflection

Set the second derivative equal to zero: \(24x = 0\). Solve for x: \(x = 0\). So, the function has a point of inflection at \(x = 0\). For function b: \(f(x) = 4x - x^2\)

Find the first derivative f'(x)

Compute the first derivative using the power rule: \(f'(x) = 4 - 2x\).

Apply the first derivative test

Find critical points by setting the first derivative equal to zero: \(4 - 2x = 0\). Solve for x: \(x = 2\). Analyze the critical point using the first derivative test: - For \(x < 2\), \(f'(1) = 2\), so the function is increasing. - For \(x > 2\), \(f'(3) = -2\), so the function is decreasing. Therefore, we have a local maximum at \(x = 2\).

Find the second derivative f''(x)

Compute the second derivative by differentiating f'(x), \(f''(x) = -2\).

Determine the points of inflection

Since the second derivative is constant, there is no point of inflection. For function c: \(f(x) = x^3\)

Find the first derivative f'(x)

Compute the first derivative using the power rule: \(f'(x) = 3x^2\).

Apply the first derivative test

Find critical points by setting the first derivative equal to zero: \(3x^2 = 0\). Solve for x: \(x = 0\). However, since the function decreases for \(x < 0\) and increases for \(x > 0\), there is no local maximum or minimum at \(x = 0\).

Find the second derivative f''(x)

Compute the second derivative by differentiating f'(x), \(f''(x) = 6x\).

Determine the points of inflection

Set the second derivative equal to zero: \(6x = 0\). Solve for x: \(x = 0\). So, the function has a point of inflection at \(x = 0\).