Geometrically the Mean Value theorem ensures that there is at least one point on the curve f(x) , whose abscissa lies in (a, b) at which the tangent is
GMVT states that for a given arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints.
When Rolle’s Theorem is verified for f(x) on [a, b] then there exists c such that
Answer is
B) c ∈ (a, b) such that f'(c) = 0.
Statement for Rolle’s Theorem :
Suppose that a function f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Then if f(a)=f(b), then there exists at least one point c in the open interval (a,b) for which f′(c)=0.
If the function f (x) = x^{2}– 8x + 12 satisfies the condition of Rolle’s Theorem on (2, 6), find the value of c such that f ‘(c) = 0
f (x) = x^{2}  8x + 12
Function satisfies the condition of Rolle's theorem for (2,6).
We need to find c for which f’(c) = 0
f’(x) = 2x – 8
f’(c) = 2c – 8 = 0
c = 4
The value of c for which Lagrange’s theorem f(x) = x in the interval [1, 1] is
For LMVT to be valid on a function in an interval, the function should be continuous and differentiable on the interval
Here,
f(x) = x , Interval : [1,1]
For h>0,
f’(0) = 1
For h<0,
f’(0) = 1
So, the LHL and RHL are unequal hence f(x) is not differentiable at x=0.
In [1,1], there does not exist any value of c for which LMVT is valid.
To verify Rolle’s Theorem which one is essential?
Rolle’s Theorem:
If a realvalued function f is continuous on a proper closed interval [a, b], differentiable on the open interval (a, b), and f (a) = f (b), then there exists at least one c in the open interval (a, b) such that
f’(c) = 0
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