Given the function: \[ y = 2x^3 - 21x^2 + 36x - 20 \] Step 1: Find the first derivative \[ \frac{dy}{dx} = 6x^2 - 42x + 36 \] Step 2: Find critical points Set \(\frac{dy}{dx} = 0\): \[ 6x^2 - 42x + 36 = 0 \] \[ x^2 - 7x + 6 = 0 \Rightarrow (x - 6)(x - 1) = 0 \] \[ x = 1 \text{ or } x = 6 \] Step 3: Apply the second derivative test Find the second derivative: \[ \frac{d^2y}{dx^2} = 12x - 42 \] - At \(x = 1\): \[ 12(1) - 42 = -30 < 0 \ (\text{Maximum at } x = 1) \] - At \(x = 6\): \[ 12(6) - 42 = 72 - 42 = 30 > 0 \ (\text{Minimum at } x = 6) \] Step 4: Find the minimum value Substitute \(x = 6\) into the original function: \[ y = 2(6)^3 - 21(6)^2 + 36(6) - 20 \] \[ = 2(216) - 21(36) + 216 - 20 \] \[ = 432 - 756 + 216 - 20 = -128 \] ✅ Final Answer: The minimum value of the function is \(-128\).