Minmini's Minimum
Find the minimum value of the function f ( x ) = x 2 − 1 2 x + 4 0 defined over all reals.
The answer is 4.
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8 solutions
Another 2 ways:
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Find the vertex x v = − 2 a b and then find f ( x v ) .
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Write the function as f ( x ) = ( x − 6 ) 2 + 4 ≥ 4 .
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Nice. I used method 2. Is there something similar to M1 for polynomials of higher degree?
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Are you asking if there's some formula for a vertex when the degree of the polynomial is higher? I really doubt.
Take a general polynomial, take its derivative, and set it equal to 0. Solve for x and see what results you get.
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@Daniel Liu – yeah, but that's not exactly what Krishna is asking for. He's asking for a way of finding the vertex in my first way, and you've shown the way Kartik Sharma used. My way consists of having an explicit formula for the vertex.
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@Mathh Mathh – I would say the vertex can be found out using the same method.
Actually, this is what we are exactly doing.
f(x) = ax^2 + bx +c
f'(x) = 2a(x) + b
f'(x) = 0
Therefore, x = -b/2a
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@Kartik Sharma – Well, I don't think about it this way, I think about it like an arithmetic mean of the roots - the vertex is exactly in the middle coordinate of the segment X 1 X 2 .
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@Mathh Mathh – Is this wrong? I derived it myself? By the way, just to make my problem famous, can you please have a look at my new problem - "Okay, but what we are given?"
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@Kartik Sharma – My 1st method asks for an explicit formula for the vertex and your derivatives are hence different from my method in a way that you'll have to solve an equation, you won't solve it just by knowing some formula.
I think differentiation is the best method once you go to higher degree polynomials..... Vertices become complicated unlike quadratics.....
Hey kartik could you just help me .? What did u do in last step? ( after getting 6)
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x = 6, hence, minimum of function is f(6) and that's what I have done.
It is substituting x by 6 only.
You could just complete the square:
x 2 − 1 2 x + 4 0 = ( x − 6 ) 2 + 4
As you can see since there is a square the smallest value of it is when x = 6 and the smallest value of the entire expression is 4 .
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Did the same !!
f ( x ) f ′ ( x ) f " ( x ) = = = x 2 − 1 2 x + 4 0 2 x − 1 2 = 0 ⟹ x = 6 2 > 0 ∴ As 6 > 2 > 0 so it is point of global minima.
Plugging in x = 6 , we get the minimum value of f ( x ) as,
f ( 6 ) = 6 2 − 1 2 ( 6 ) + 4 0 = 3 6 − 7 2 + 4 0 = 4
let f(x)=x^{2}-12x+40
then f(x)=(x-6)^{2}+4
as no square is negative, any square's minimum value is 0
hence x=6 for minimum value
and f( 6)=4 which gives the minimum value of f(x) is 4
One can also write the given function as being equal to ( X - 6 )^2 + 4. Clearly, the minimum value of the function is 4 and is acheived at X = 6.
derivative must be zero ,,, and will got x = 6 thn put x = 6 in the function
just solve the vertex. and the value of y in the vertex is the minimum value of the function or use this formula: (4ac - b^2)/4a
derivative of f(x) = 2x - 12
Now, for it to be minimum, derivative must be 0.
Hence, 2x-12 = 0
x= 6
Therefore, (6)^2 - 12(6) + 40 = 76 -72 = 4