# Global (Absolute) Maxima or Minima

## Global (Absolute) Maxima or Minima

Global (Absolute) Maxima or Minima in [a, b]:

Step 1:  Find out all the critical points of f(x) in (a, b). Let c₁, c₂, …, cn be the different critical points.

Step 2: Find the value of the function at these critical points and also at the end points of the domain. Let the values are f(c₁), f(c₂), …, f(cn).

Step 3: Find M₁ = Max {f(a), f(c₁), f(c₂), …, f(cn), f(b)} and M₂ = Min {f(a), f(c₁), f(c₂), …, f(cn), f(b)}

Now, M₁ is the maximum value of f(x) in [a, b], so M₁ is absolute maximum and M₂ is the minimum value of f(x) in [a, b], so M₂ is absolute minimum.

Let y = f(x) be the function defined on [a, b] in the graph, then

(i) f(x) has local maximum values at x = a₁, a₃, a₅, a₇

(ii) f(x) has local minimum values at x = a₂, a₄, a₆, a₈

(iii) The absolute maximum value of the function is f(a₇) and absolute minimum value is f(a).

Note:

• Between two local maximum values, there is a local minimum values and vice – versa.
• A local minimum value may be greater than a local maximum value.

Global (Absolute) Maxima or Minima in (a, b):

To find the absolute maxima and minima in (a, b) step1 and step2 are same. Now

Step 4: Find M₁ = Max {f(c₁), f(c₂), ….., f(cn)} and M₂ = Min {f(c₁), f(c₂),……, f(cn)}.

Now,

$$\underset{\begin{smallmatrix}\ \ \ \ x\to {{a}^{+}} \ (or)x\to {{b}^{-}} \end{smallmatrix}}{\mathop{\lim }}\,f(x)>{{M}{1}}\ \ (or)\ \underset{\begin{smallmatrix} \ \ \ \ \ x\to {{a}^{+}} \ (or)x\to {{b}^{-}} \end{smallmatrix}}{\mathop{\lim}}\,f(x)<{{M}{2}}$$.

Then f(x) would not have absolute maximum or obsolete minimum in (a, b) and if

$$\underset{x\to {{a}^{+}}\ and\ x\to {{b}^{-}}}{\mathop{\lim }}\,f(x)<{{M}_{1}}\ \ \ and\ \ \underset{x\to {{a}^{+}}\ and\ x\to {{b}^{-}}}{\mathop{\lim }}\,f(x)>{{M}_{2}}$$.

Then M₁ and M₂ would respectively the absolute maximum and absolute minimum of f(x) in (a, b).