Question

Verify Lagrange's mean value theorem for

f(x): (4x-1)^-1 in the interval [1,4]

Answer 1

Toolbox:

• Lagrange's Mean Value Theorem :
• Let f(x)be a real valued function that satisfies the following conditions.
• (i) f(x) is continuous on the closed interval [a,b][a,b]
• (ii) f(x) is differentiable in the open interval (a,b)

• (iii) if f(a)=f(b)
• Then there exists atleast one value c∈(a,b)such that f′(c)=0 where,
• f′(c)=f(b)−f(a)/b−a

Step 1:
Given :f(x)=(4x-1)^-1 in the interval [1,4]
We know that a polynomial function is continuous everywhere and also differentiable.
So f(x) being a polynomial is continuous and differentiable on (1,4).
So there must exist at least one real number c∈(1,4)such that
f′(c)=f(4)−f(1)/4−1
Step 2:

f(x)=(4x-1)^-1=1/(4x-1)
f(4)=(4×4-1)^-1=1/15
f(1)=(4×1-1)^-1=1/3

f′(x)=-1×4/(4x-1)^2
=-4/(4x-1)^2

f′(c)=-4/(4c-1)^2

According to the condition ,
f'(c)=f(b)-f(a)/b-a


c∈(1,4)
Hence Lagrange's Mean Value theorem is verified.