what is the minimum value of 7^(x^2-4x+7)

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what is the minimum value of 7^(x^2-4x+7)

Mathematics
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\[\Large 7^{x^2-4x+7}\]Like this?
yes

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Other answers:

Minimum value should be there where dy/dx = 0.
what do you mean by the "d"?
In other words minimum value occurs where the derivative of the function is equal to zero.
isn't derivatives calculus? i haven't learned that yet and i don't think we're expected to know it to solve this problem
http://www.mememaker.net/static/images/templates/14288.jpg
haha i was thinking about it and wouldn't it be one over infinity? because if x^2-4x+7 was negative infinity than 7^(x^2-4x+7) would be 1/infinity
Hint: 7^x is an increasing function. So if we minimize x, 7^x is also a minimum.
x^2-4x+7 cannot be at -inf. It is a parabola concave upwards.!
And are we going to try that by trial and error? @mathmate
No, we need to simply find the vertex of the parabola to find the minimum value of the parabola.
Howw??
Good point!] @livelaughlilz can you take it from here?
Completing the squares will do the job.
yeah the vertex of the parabola is (2,3)
(x-2)^2=y-3
x^2-4x+7=(x-2)^2+3 so the vertex is at (2,3)
http://i1.kym-cdn.com/entries/icons/original/000/009/993/tumblr_m0wb2xz9Yh1r08e3p.jpg
Good job livelaughlilz! Speed+accuracy!
I like his movies! :)
oh so the answer is 343. the lowest value of x is 2, and if you plug it in you get 7^3=343
7^2 = ???
@mathmate : haha, nice. My brother just love his movies. i don't. :/
Nevermind, I was out of my mind. Yes, 7^3=343
You were thinking about Jackie Chan. :D
As a sidenote @mathmate, you can find the x-coordinate of the vertex using the formula \[v_x=\frac{-b}{2a}\]
Thank you for the defence, even though it was a lame one! Good thinking! :)
I simply love this method. Satellite73 taught me this. -b/2a
to add onto @KingGeorge the y-coordinate of the vertex is -[(b^2-4ac)/(4a)]
Thank you @KingGeorge, my memory is limited and sometimes defective. But thanks for the shortcut anyway!
you can figure all that out by completing the square of y=ax^2+bx+c
and making y=ax^2+bx+c into vertex form
Actually I know the -b/2a part, that's how I get to do the completing the square. It's just I never managed to memorize the -[(b^2-4ac)/(4a)] part.
well b^2-4ac is also the discriminant...
@livelaughlilz is a boss at this.
Perhaps from this time on I will remember. Practice makes perfect! :)
Actually what I do is finding it by c-(b/2a)^2, which is exactly what we do when we complete the square.
oh
i'm going to close this question now. thanks everyone!
See you all!

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