Applications of Derivatives

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New answer posted

a month ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

f (2)=8, f' (2)=5, f' (x) ≥ 1, f' (x) ≥ 4, ∀x ∈ (1,6)
Using LMVT
f' (x) = (f' (5) - f' (2)/ (5-2) ≥ 4 ⇒ f' (5) ≥ 17
f' (x) = (f (5) - f (2)/ (5-2) ≥ 1 ⇒ f (5) ≥ 11
Therefore f' (5) + f (5) ≥ 28

New question posted

a month ago

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New answer posted

a month ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

f (x) = (3x - 7)x²/³
⇒ f (x) = 3x? /³ - 7x²/³
⇒ f' (x) = 5x²/³ - 14/ (3x¹/³)
= (15x - 14) / (3x¹/³) > 0


∴ f' (x) > 0 ∀x ∈ (-∞, 0) U (14/15, ∞)

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

f(3)=f(4) ⇒ α=12
f'(x) = (x²-12)/(x(x²+12))
∴ f'(c)=0 ⇒ c=√12
∴ f''(c) = 1/12

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

0.50
 y = x²-3x+2, x+y=a, x-y=b
x?=2 x?=1
y? = 4-6+2 = 0 y?=0
(2,0)
(1,0)
b=2
a=1
∴ a/b = 1/2 = 0.5

New answer posted

a month ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

e? y'x? + 4x³e? + 2y' / (2√ (y+1) = 0 at (1,0)
y' + 4 + y' = 0 ⇒ y' = -2
equation of tangent at (1,0) is 2x + y - 2 = 0
So option (C) is correct.

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