Find a value of k that gives f(x)=x²+kx+2 a local minimum value of 1.

Question

I do not not what to do.

Answer 1

try completing the square.

x^2+kx+2
= x^2 + kx + (k/2)^2 + 2 -(k/2)^2
= (x + k/2) + 2 - k^2/4

Now you have the vertex form of the parabola, with the vertex at x = -k/2. At that point, the minimum is

2 - k^2/4

So, now just set

2 - k^2/4 = 1
k^2/4 = 1
k^2 = 4
k = ±2

f(x) = x^2+2x+2 = (x+1)^2 + 1
f(x) = x^2-2x+2 = (x-1)^2 + 1

Both have a minimum value of 1 at x = ±1

Answer 2

oh. right. you want to use calculus.

f' = 2x+k
f'=0 at x = -k/2

f(-k/2) = k^2/4 + k(-k/2) + 2
= k^2/4 - k^2/2 + 2
= -k^2/4 + 2

we want the minimum to be 1, so

-k^2/4 + 2 = 1
k^2/4 = 1
k^2 = 4
k = ±2
as above