Lagrange Theorem

swisshome

New member
Joined
May 17, 2019
Messages
1
I've got the following exercise to solve. Using Lagrange Theorem for the function f(x)=\(\displaystyle sqrt(x)\) for a=1, and b=9. Find the value of x=c.
Now I checked the conditions.
1) f(x) is continuous on [1,9]
2) f(x) is differentiable on (1,9)
Hence, f'(c)=1/4
Now how to find c if f'(c)=1/4?
f(c)=1/4x+C if I'm right
 

LCKurtz

Junior Member
Joined
May 3, 2019
Messages
77
I've got the following exercise to solve. Using Lagrange Theorem for the function f(x)=\(\displaystyle sqrt(x)\) for a=1, and b=9. Find the value of x=c.
Now I checked the conditions.
1) f(x) is continuous on [1,9]
2) f(x) is differentiable on (1,9)
Hence, f'(c)=1/4
Now how to find c if f'(c)=1/4?
f(c)=1/4x+C if I'm right
Your last line is incorrect. You have \(\displaystyle f(x) = \sqrt x\). What is \(\displaystyle f'(x)\)? Set \(\displaystyle f'(x)=4\) and solve for \(\displaystyle x\) to get \(\displaystyle c\).
 
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