Calculus!!?

Question

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

Ok the answer for this question is k = 2 or k = - 2

(I got k = -2 but I dont see how they got k = 2)

Someone save me plz....

Answer 1

f(x) = x^2 + kx + 2

We obtain the local minimum by taking the derivative and then making it 0.

f'(x) = 2x + k

Make f'(x) = 0,

0 = 2x + k

Solve for x.

-k = 2x

x = -k/2

Therefore, if there exists a minimum at x = -k/2, it follows that
f(-k/2) = 1. But, by the definition of the function,

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

f(-k/2) = (k^2)/4 - (2k^2)/4 + 8/4

f(-k/2) = (k^2 - 2k^2 + 8)/4
f(-k/2) = (-k^2 + 8)/4

But f(-k/2) = 1, so

(-k^2 + 8)/4 = 1
-k^2 + 8 = 4
-k^2 = -4
k^2 = 4

Take the square root of both sides; remember to include "plus or minus" when doing so, as it is part of the square root property.

k = +/- 2

Answer 2

Here's how:
f(x) = x² + kx + 2 ⇒ f '(x) = 2x + k

This is zero when 2x + k = 0 ⇒ x = -½k

Sub this into the original formula for f, and use the fact that we must have f(-½k) = 1

(-½k)² + k(-½k) + 2 = 1
¼k² - ½k² + 1 = 0
this leads to k² = 4, so k = ±2

Answer 3

f(x) = x^2 + kx + 2

f'(x) = 2x + k --> this tells you what value of X makes f(x) a max/min when set equal to zero.

so when x = -k/2 , f(x) is a min (in this case 1)

1 = (-k/2)^2 +k(-k/2) + 2

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

-1 = -k^2/4

k^2 = 4

k= 2 or k = -2

(remember sqrt always returns plus and minus values)

Answer 4

the graph of this function is a parabola opening up.
Minimum at the vertex. Coordinates of the vertex are:
V(-b/2a; -discriminant/4a)

-discriminant/4a = 1
discriminant=b^2-4ac=k^2-8
(-k^2+8)/4=1
-k^2+8=4
-k^2=-4
k^2=4
k=2 or k=-2

The minimum value of this function will be 1 (y=1) for k=2 or -2