If f(x) = 16x^2 + 19x - 32 what is the minimum value of f(x).

### 1 Answer | Add Yours

The function f(x) = 16x^2 + 19x - 32. At the point where the value of f(x) is the least f'(x) = 0. In addition, for the solution x = a of f'(x) = 0, the value of f''(a) should be positive.

f(x) = 16x^2 + 19x - 32

f'(x) = 32x + 19

f''(x) = 32

f'(x) = 0

=> x = -19/32

The value of f''(x) is positive for all values of x.

At x = -19/32, f(x) = -2406/64

**The minimum value of f(x) is -2406/64**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes