Tool to determine the minimum value of a function: the minimal value that can take a function. It is a global minimum and not a local minimum.
Minimum of a Function - dCode
Tag(s) : Functions
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For any function $ f $ defined on an interval $ I $ and $ m $ a real number belonging to $ I $, if $ f(x) <= f(m) $ on the interval $ I $ then $ f $ reaches its minimum in $ x=m $ over $ I $. In that case, $ f(m) $ is the minimum value of the function, reached when $ x=m $.
The minimum of a function is always defined with an interval (which may be the whole domain of definition of the function).
Finding the minimum of a function $ f $, is equivalent to calculate $ f(m) $. To find $ m $, use the derivative of the function. The minimum value of a function is found when its derivative is null and changes of sign, from negative to positive.
Example: $ f(x) = x^2 $ defined over $ \mathbb{R} $, its derivative is $ f'(x) = 2x $, that is equal to zero in $ x = 0 $ because $ f'(x) = 0 \iff 2x = 0 \iff x=0 $. The derivative goes from negative to positive in $ x = 0 $ so the function has a minimum in $ x=0 $, $ f(x=0) = 0 $ and $ f(x) >= 0 $ over $ \mathbb{R} $.
Add one or more conditions indicating the interval constraints for each variable.
Example: Find the minimum of $ \sin{x} $ for $ 0 < x < \pi $
Indicate several equations with the operator logical AND && to separate the equations
An extremum is an extreme value of a function, this value can be maximum (the maximum value of the function) or minimum (the minimum value of the function).
The minorant is any value lower than or equal to the minimum value reached by the function.
A constant function $ f (x) = c $ is a line that always equals $ c $, so its minimum is $ c $, reached for any value of $ x $
An line/affine function $ f (x) = ax + b $ always has for minimum $ -\infty $
— If $ a < 0 $, the minimum of $ f $ is $ -\infty $ when $ x $ tends to $ +\infty $
— If $ a > 0 $, the minimum of $ f $ is $ -\infty $ when $ x $ tends to $ -\infty $
A quadratic polynomial function of the form $ f(x) = ax^2+bx+c $ then
— If $ a > 0 $, the minimum of $ f $ is $ (-b^2 + 4 a c)/(4 a) $ reached when $ x = -\frac{b}{2a} $
— If $ a < 0 $, the minimum of $ f $ is $ +\infty $ when $ x $ tends to $ +\infty $
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