Find the maximum and minimum values of the function $f(x) = 2x^3 - 15x^2 + 36x + 1$.

Question

Accepted Answer

**Step 1: Find the first derivative $f'(x)$**
$$f'(x) = 6x^2 - 30x + 36$$

**Step 2: Find critical points by setting $f'(x) = 0$**
$$6(x^2 - 5x + 6) = 0 \implies (x - 2)(x - 3) = 0$$
Critical points are $x = 2$ and $x = 3$.

**Step 3: Find the second derivative $f''(x)$**
$$f''(x) = 12x - 30$$

**Step 4: Test critical points**
- At $x = 2$:
  $$f''(2) = 12(2) - 30 = -6 < 0 \implies 	ext{Local Maximum at } x = 2$$
  $f(2) = 2(8) - 15(4) + 36(2) + 1 = 16 - 60 + 72 + 1 = 29$

- At $x = 3$:
  $$f''(3) = 12(3) - 30 = 6 > 0 \implies 	ext{Local Minimum at } x = 3$$
  $f(3) = 2(27) - 15(9) + 36(3) + 1 = 54 - 135 + 108 + 1 = 28$

Maximum value is **29** and minimum value is **28**.

Find the maximum and minimum values of the function $f(x) = 2x^3 - 15x^2 + 36x + 1$. - DDCET Q&A

Find the maximum and minimum values of the function $f(x) = 2x^3 - 15x^2 + 36x + 1$.

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