livelaughlilz
what is the minimum value of 7^(x^24x+7)



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saifoo.khan
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\[\Large 7^{x^24x+7}\]Like this?

livelaughlilz
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yes

saifoo.khan
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@mathmate @UnkleRhaukus

saifoo.khan
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Minimum value should be there where dy/dx = 0.

livelaughlilz
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what do you mean by the "d"?

saifoo.khan
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In other words minimum value occurs where the derivative of the function is equal to zero.

livelaughlilz
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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


livelaughlilz
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haha
i was thinking about it and wouldn't it be one over infinity? because if x^24x+7 was negative infinity than 7^(x^24x+7) would be 1/infinity

mathmate
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Hint: 7^x is an increasing function. So if we minimize x, 7^x is also a minimum.

mathmate
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x^24x+7 cannot be at inf. It is a parabola concave upwards.!

saifoo.khan
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And are we going to try that by trial and error? @mathmate

KingGeorge
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No, we need to simply find the vertex of the parabola to find the minimum value of the parabola.

saifoo.khan
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Howw??

mathmate
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Good point!]
@livelaughlilz can you take it from here?

mathmate
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Completing the squares will do the job.

livelaughlilz
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yeah the vertex of the parabola is (2,3)

livelaughlilz
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(x2)^2=y3

mathmate
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x^24x+7=(x2)^2+3
so the vertex is at (2,3)


mathmate
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Good job livelaughlilz! Speed+accuracy!

mathmate
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I like his movies! :)

livelaughlilz
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oh so the answer is 343. the lowest value of x is 2, and if you plug it in you get 7^3=343

saifoo.khan
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Yes^ @livelaughlilz

mathmate
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7^2 = ???

saifoo.khan
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@mathmate : haha, nice. My brother just love his movies. i don't. :/

mathmate
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Nevermind, I was out of my mind. Yes, 7^3=343

saifoo.khan
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You were thinking about Jackie Chan. :D

KingGeorge
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As a sidenote @mathmate, you can find the xcoordinate of the vertex using the formula \[v_x=\frac{b}{2a}\]

mathmate
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Thank you for the defence, even though it was a lame one! Good thinking! :)

saifoo.khan
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I simply love this method. Satellite73 taught me this. b/2a

livelaughlilz
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to add onto @KingGeorge the ycoordinate of the vertex is [(b^24ac)/(4a)]

mathmate
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Thank you @KingGeorge, my memory is limited and sometimes defective. But thanks for the shortcut anyway!

livelaughlilz
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you can figure all that out by completing the square of y=ax^2+bx+c

livelaughlilz
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and making y=ax^2+bx+c into vertex form

mathmate
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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^24ac)/(4a)] part.

livelaughlilz
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well b^24ac is also the discriminant...

saifoo.khan
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@livelaughlilz is a boss at this.

mathmate
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Perhaps from this time on I will remember. Practice makes perfect! :)

mathmate
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Actually what I do is finding it by c(b/2a)^2, which is exactly what we do when we complete the square.

livelaughlilz
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oh

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

mathmate
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See you all!