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).