Answer :

**Given:** f(x) = x^{4} - 8x^{3} + 22x^{2} - 24x + 21

Finding f ^{‘}(x)

f ^{‘}(x) = 4x^{3} - 24x^{2} + 44x - 24

**formula**:

f ^{‘}(x) = 4(x^{3} - 6x^{2} + 11x - 6)

f ^{‘}(x) = 4(x - 1) (x - 2) (x - 3)

for finding the maxima or minima f ^{‘}(x) = 0 at that point

f ^{‘}(x) = 0 at x = 1, x = 2 and x = 3

the intervals are ( - ∞, 1) (1, 2) (2, 3) (3, ∞)

since f ^{‘}(x) > 0 in (1, 2) and (3, ∞)

therefore f(x) is strictly increasing in (1, 2) and (3, ∞) and strictly decreasing in ( - ∞, 1) and (2, 3).

**OR**

**Given:** f(x) = sec x + log cos^{2}x

Finding f ^{‘}(x)

f ^{‘}(x) = sec x tan x - 2tan x

**Formula**:

f ^{‘}(x) = tan x (sec x - 2)

for finding the maxima or minima f ^{‘}(x) = 0 at that point

f ^{‘}(x) = 0 at tan x = 0 or sec x = 2

therefore, solution is x = π or

now finding f” (x) to check whether the point is maxima or minima

f” (x) = sec x tan^{2}x + (sec x - 2) sec^{2}x

**Formula:**

substituting the points

which is positive hence it is minimum

which is negative hence it is maximum

which is positive hence it is minimum

Therefore, we can conclude that

Maximum value

Minimum value

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