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Find k so that f(x) = x^2 + kx +2 has a local min at y = 1
- Thread starter fred2028
- Start date
Re: Find k
You can use \(\displaystyle x=\frac{-b}{2a}\), which gives the x-coordinate of the vertex of your parabola.
You get \(\displaystyle x=\frac{-k}{2}\)
Plug this back into your equation, set it to 1 and solve for k.
You can use \(\displaystyle x=\frac{-b}{2a}\), which gives the x-coordinate of the vertex of your parabola.
You get \(\displaystyle x=\frac{-k}{2}\)
Plug this back into your equation, set it to 1 and solve for k.
Re: Find k
You almost have it. Solve the derivative you have for k and sub into the equation.
You have 2x+k=0
k=-2x
\(\displaystyle x^{2}+(-2x)x+2=1\)
Solve for x.
but surely there's a calculus way?
You almost have it. Solve the derivative you have for k and sub into the equation.
You have 2x+k=0
k=-2x
\(\displaystyle x^{2}+(-2x)x+2=1\)
Solve for x.
Re: Find k
So with that I getgalactus said:You can use \(\displaystyle x=\frac{-b}{2a}\), which gives the x-coordinate of the vertex of your parabola.
You get \(\displaystyle x=\frac{-k}{2}\)
Plug this back into your equation, set it to 1 and solve for k.
But the answer is +/- 2(-k/2)^2 + k(-k/2) + 2 = 1
(k^2)/4 - (k^2)/2 + 1 = 0
k^2 - 2k^2 + 1 = 0
-k^2 + 1 = 0
k^2 = 1
k = +/- 1
Re: Find k
fred2028 said:So with that I get
But the answer is +/- 2(-k/2)^2 + k(-k/2) + 2 = 1
(k^2)/4 - (k^2)/2 + 1 = 0
k^2 - 2k^2 + 1 = 0 Seems like you multiplied both sides by 4. You forgot to multiply + 1 by it.
-k^2 + 1 = 0
k^2 = 1
k = +/- 1
Re: Find k
Edit: OK that other method which you corrected me on works out fine ...
That still gives me +/- 1 for some reason, and the back of the book says +/- 2. I graphed it, and it should be +/- 2, since +/- 1 gives a local min of roughly 1.8.galactus said:but surely there's a calculus way?
You almost have it. Solve the derivative you have for k and sub into the equation.
You have 2x+k=0
k=-2x
\(\displaystyle x^{2}+(-2x)x+2=1\)
Solve for x.
Edit: OK that other method which you corrected me on works out fine ...
Re: Find k
The answer is +/-2. Recheck your algebra.
\(\displaystyle (\frac{-k}{2})^{2}+(\frac{-k}{2})k+2 = 2-\frac{k^{2}}{4}\)
\(\displaystyle 2-\frac{k^{2}}{4}=1\)
\(\displaystyle \frac{k^{2}}{4}=1\)
\(\displaystyle \frac{k^{2}}=4\)
\(\displaystyle k=\pm\sqrt{2}\)
And yes, x = +/-1. Plug that in k=-2x and get k=+/-2
The answer is +/-2. Recheck your algebra.
\(\displaystyle (\frac{-k}{2})^{2}+(\frac{-k}{2})k+2 = 2-\frac{k^{2}}{4}\)
\(\displaystyle 2-\frac{k^{2}}{4}=1\)
\(\displaystyle \frac{k^{2}}{4}=1\)
\(\displaystyle \frac{k^{2}}=4\)
\(\displaystyle k=\pm\sqrt{2}\)
And yes, x = +/-1. Plug that in k=-2x and get k=+/-2
Re: Find k
They both work fine. One gives x and the other gives k.
OK that other method which you corrected me on works out fine ...
They both work fine. One gives x and the other gives k.
Dr. Flim-Flam
Junior Member
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- Oct 10, 2007
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f(x) = x^2 + kx +2; f '(x) = 2x + k = 0, x =-k/2
f(-k/2) = 1 = k^2/4 -k^2/2 +2. k= 2, k = -2.
When k = 2, f(x) = x^2 + 2x + 2, f '(x) = 2x +2 =0, x = -1, f(-1) = 1
When k = -2, f(x) = x^2 -2x +2, f' '(x) = 2x-2 = 0, x = 1, f(1) = 1.
Hence the equations x^2 + 2x +2 and x^2 -2x + 2 have a local min at y = 1.
f(-k/2) = 1 = k^2/4 -k^2/2 +2. k= 2, k = -2.
When k = 2, f(x) = x^2 + 2x + 2, f '(x) = 2x +2 =0, x = -1, f(-1) = 1
When k = -2, f(x) = x^2 -2x +2, f' '(x) = 2x-2 = 0, x = 1, f(1) = 1.
Hence the equations x^2 + 2x +2 and x^2 -2x + 2 have a local min at y = 1.