Application of Derivatives Test - 8

Given f(x) = x² - 8x + 17

⇒ f'(x) = 2x - 8

For minimum valye of f(x) we have f'(x) = 0

⇒ 2x - 8 = 0 ⇒⇒ x = 8282 = 4

Now, f''(x) = 2 > 0, hence the minimum of f(x) exist at x = 4 and minimum value = f(4) = 1.  

Let f(x) = alogx + bx² + x

f′(x) = a.1/x + 2bx + 1

For maximum and minimum value of f(x) we have f'(x) = 0

Therefore, at x = -1 and x = 2 we have 2bx² + x + a = 0

i.e, a + 2b = 1....(i) and a + 8b = 2....(ii)

(ii) - (i ) gives b = −1/2

Now , from (i) we get a = 2

⇒ a = 2, b = −1/2

we have, f(x)=sinx,then, f'(x)=cosx

For stationary points,we must have  f'(x)=0 ⇒ cosx = 0 ⇒ x = 3π/2 [∵ x ∈ [π,2π] ]

Now ,f(π) = sinπ = 0, f(3π/2) = sin 3π/2 = -1 and f(2π) = sin2π = 0

Hence,the maximum value of f(x) is 0.

Given, f(x) = x - sinx

f'(x) = 1 - cosx

For maximum or minimum values of f(x), we have f'(x) = 0

Now, f'(x) = 0  ⇒1 − cosx = 0 ⇒ cosx = 1 ⇒ x = 0

Now, f(0) = 0 -sin 0 = π − 0 = π

Hence, minimum value of f(x) is 0.

f(x) = x + cosx 

f'(x) = 1 - sinx

For maximum and minimum values of f(x) , we have f'(x) = 0

Now, f'(x) = 0 ⇒ 1 – sinx = 0 ⇒ x = π/2

Hence, maximum value of f(x) is f(π/2) = π/2

Given f(x) = x³

f'(x) = 3x²

For point of inflexion, we have f'(x) = 0

f′(x) = 0 ⇒ 3x² = 0 ⇒ x = 0

Hence, f(x) has a point of inflexion at x = 0.

But x = 0 is not a local extremum as we cannot find an interval I around x = 0 such that f(0) ≥ f(x) or f(0) ≤ f(x) for all xϵI