Question:
If \(f:[-5,5]→R\) is a differentiable function and if \(f'(x)\)does not vanish anywhere, then prove that \(f(-5)≠f(5)\)
If \(f:[-5,5]→R\) is a differentiable function and if \(f'(x)\)does not vanish anywhere, then prove that \(f(-5)≠f(5)\)
Updated On: Sep 13, 2023
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Solution and Explanation
It is given that \(f:[-5,5]→R\) is a differentiable function.
Since every differentiable function is a continuous function, we obtain
(a) \(f\) is continuous on \([−5,5]. \)
(b) \(f\) is differentiable on \((−5,5).\)
Therefore, by the Mean Value Theorem,there exists \(c∈(−5,5)\) such that
\(f'(c)=\frac{f(5)-f(-5)}{5-(-5)}\)
\(⇒10f'(c)=f(5)-f(-5)\)
It is also given that \(f'(x)\) does not vanish anywhere
\(∴f'(c)≠0\)
\(⇒10f'(c)≠0\)
\(⇒f(5)-f(-5)≠0\)
\(⇒f(-5)≠f(5)\)
Hence, proved.
Since every differentiable function is a continuous function, we obtain
(a) \(f\) is continuous on \([−5,5]. \)
(b) \(f\) is differentiable on \((−5,5).\)
Therefore, by the Mean Value Theorem,there exists \(c∈(−5,5)\) such that
\(f'(c)=\frac{f(5)-f(-5)}{5-(-5)}\)
\(⇒10f'(c)=f(5)-f(-5)\)
It is also given that \(f'(x)\) does not vanish anywhere
\(∴f'(c)≠0\)
\(⇒10f'(c)≠0\)
\(⇒f(5)-f(-5)≠0\)
\(⇒f(-5)≠f(5)\)
Hence, proved.
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Concepts Used:
Mean Value Theorem
The theorem states that for a curve f(x) passing through two given points (a, f(a)), (b, f(b)), there is at least one point (c, f(c)) on the curve where the tangent is parallel to the secant passing via the two given points is the mean value theorem.
The mean value theorem is derived herein calculus for a function f(x): [a, b] → R, such that the function is continuous and differentiable across an interval.
- The function f(x) = continuous across the interval [a, b].
- The function f(x) = differentiable across the interval (a, b).
- A point c exists in (a, b) such that f'(c) = [ f(b) - f(a) ] / (b - a)