Find the minimum value of f(x)=2x^2-6x+5

Question

queelius · Answer

Hint: Find the first derivative of f(x), which we will denote f'(x). Now, we want to know two things: (1) where does f'(x) have a slope of 0? At any such points, for it to be a minimum, it must be concave up (which means f(x) is increasing at that point).

So, solve for f'(x) = 0, and f''(x) at that point(s) is positive.

anonymous · Answer

Is it 37/2? @queelius

queelius · Answer

No, that's not correct.

f'(x) = 4x - 6. What's the next step?

queelius · Answer

Btw, I just realized you may not know Calculus. In this case, there are other ways to solve this question, e.g., by finding the vertex.

anonymous · Answer

@queelius Is it -3/2?
Sorry, it takes me a while to get this..

queelius · Answer

4x - 6 = 0; x = 3/2; f(3/2) = 2(3/2)^2-6(3/2)+5 = 2(9/4)-18/2+5 = 9/2-18/2+10/2 = 1/2

queelius · Answer

We know 1/2 is a local minimum because f(x) is increasing everywhere (f''(x) = 4, which is positive at any value of x), so whenever f(x) has a slope of zero, we know that it must be increasing (concave up) therefore it's a local minimum. In fact, since this is the only point where the slope is 0, it must be a global minimum.