Question

The local minimum value of the real function \[ f(x) = 2x^3 - 21x^2 + 36x - 20 \] is __________ (in integer).

Hint:

The second derivative test helps determine whether a critical point is a local maximum or minimum.

Answer 1

Correct Answer: -128

To find the local minimum value of the function, we need to first find the critical points by taking the derivative of the function and setting it equal to zero.
The derivative of \( f(x) \) is: \[ f'(x) = 6x^2 - 42x + 36 \] Set the derivative equal to zero to find the critical points: \[ 6x^2 - 42x + 36 = 0 \] Divide the equation by 6: \[ x^2 - 7x + 6 = 0 \] Factoring the quadratic equation: \[ (x - 1)(x - 6) = 0 \] Thus, the critical points are \( x = 1 \) and \( x = 6 \).
Now, we need to check which one is a local minimum by using the second derivative test. The second derivative of \( f(x) \) is: \[ f''(x) = 12x - 42 \] Evaluate the second derivative at the critical points: For \( x = 1 \): \[ f''(1) = 12(1) - 42 = -30 \] Since \( f''(1)<0 \), \( x = 1 \) is a local maximum.
For \( x = 6 \): \[ f''(6) = 12(6) - 42 = 30 \] Since \( f''(6)>0 \), \( x = 6 \) is a local minimum.
Now, calculate the value of the function at \( x = 6 \): \[ f(6) = 2(6)^3 - 21(6)^2 + 36(6) - 20 = 432 - 756 + 216 - 20 = -128 \] Thus, the local minimum value of the function is \( -128 \).
Final Answer: -128