Question

Find the local maximum value and local minimum value (whichever exists) for the function \[ f(x) = 4x^2 + \frac{1}{x}, \quad (x \neq 0). \]

Hint:

Use the first derivative test to find critical points and the second derivative test to classify them as local maxima or minima.

Answer 1

To find the critical points, differentiate \( f(x) \) with respect to \( x \): \[ f'(x) = \frac{d}{dx} \left( 4x^2 + \frac{1}{x} \right) = 8x - \frac{1}{x^2}. \] Set \( f'(x) = 0 \) to find the critical points: \[ 8x - \frac{1}{x^2} = 0 \quad \Rightarrow \quad 8x^3 = 1 \quad \Rightarrow \quad x = \frac{1}{2}. \] To classify the critical point, compute the second derivative: \[ f''(x) = \frac{d}{dx} \left( 8x - \frac{1}{x^2} \right) = 8 + \frac{2}{x^3}. \] Substitute \( x = \frac{1}{2} \): \[ f''\left(\frac{1}{2}\right) = 8 + \frac{2}{\left(\frac{1}{2}\right)^3} = 8 + 16 = 24>0. \] Since \( f''(x)>0 \), the function has a local minimum at \( x = \frac{1}{2} \). Find the value of \( f(x) \) at \( x = \frac{1}{2} \): \[ f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^2 + \frac{1}{\frac{1}{2}} = 4\left(\frac{1}{4}\right) + 2 = 1 + 2 = 3. \] Final Answer: The local minimum value of \( f(x) \) is \( 3 \).