Question

What are the critical points of the function f(x) ?

Answer 1

Step 1: Differentiate the function:
\( f(x) = \frac{1}{3}x^3 - 4x^2 + 15x + 2 \)
So, the derivative is:
\( f'(x) = \frac{d}{dx} \left( \frac{1}{3}x^3 - 4x^2 + 15x + 2 \right) = x^2 - 8x + 15 \)
Step 2: Set \( f'(x) = 0 \) to find critical points:
\( x^2 - 8x + 15 = 0 \)
Solve using quadratic formula:
\( x = \frac{8 \pm \sqrt{64 - 60}}{2} = \frac{8 \pm 2}{2} \)
\( \Rightarrow x = 3, 5 \)

Answer 2

Step 3: Find the second derivative:
\( f''(x) = 2x - 8 \)
Step 4: Evaluate second derivative at critical points:
\( f''(3) = 2(3) - 8 = -2<0 \Rightarrow x = 3 \) is a maximum
\( f''(5) = 2(5) - 8 = 2>0 \Rightarrow x = 5 \) is a minimum
Step 5: Find the minimum value of the function:
\( f(5) = \frac{1}{3}(125) - 4(25) + 15(5) + 2 \)
\( = \frac{125}{3} - 100 + 75 + 2 \)
\( = \frac{125}{3} - 23 = \frac{125 - 69}{3} = \frac{56}{3} \approx 18.67 \)
Minimum value: ≈ 18.67