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

yes

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.

http://www.mememaker.net/static/images/templates/14288.jpg

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

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)]

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

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! :)

oh

i'm going to close this question now. thanks everyone!

See you all!