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Prove that the logarithmic function is strictly increasing on (0, ∞).

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Prove that the logarithmic function is strictly increasing on (0,∞).(0,∞).

cbse class12 bookproblem ch6 sec2 q10 p206 sec-aeasy math

  

 

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asked Nov 23, 2012 by thanvigandhi_1 
edited Jul 9, 2013 by sreemathi.v



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Toolbox:A function f(x)f(x) is said to be a strictly increasing function on (a,b)(a,b) if x1<x2⇒f(x1)<f(x2)x1<x2⇒f(x1)<f(x2) for all x1,x2∈(a,b)x1,x2∈(a,b)If x1<x2⇒f(x1)>f(x2)x1<x2⇒f(x1)>f(x2) for all x1,x2∈(a,b)x1,x2∈(a,b) then f(x)f(x) is said to be strictly decreasing on (a,b)(a,b)A function f(x)f(x) is said to be increasing on [a,b][a,b] if it is increasing (decreasing) on (a,b)(a,b)and it is increasing (decreasing) at x=ax=a and x=bx=b.The necessary sufficient condition for a differentiable function defined on (a,b)(a,b) to be strictly increasing on (a,b)(a,b) is that f′(x)>0f′(x)>0for all x∈(a,b)x∈(a,b)The necessary sufficient condition for a differentiable function defined on (a,b)(a,b) to be strictly decreasing on (a,b)(a,b) is that f′(x)<0f′(x)<0for all x∈(a,b)x∈(a,b)

Step 1:

Let f(x)=logxf(x)=log⁡x

Differentiating w.r.t xx we get,

f′(x)=1xf′(x)=1x

Step 2:

Clearly when f′(x)>0f′(x)>0

⇒1x⇒1x>0>0 when x>0x>0

Therefore f′(x)>0f′(x)>0

Hence f(x)f(x) is an increasing function for x>0x>0

(i.e) f(x)f(x) is increasing function whenever it is defined.