Question:
Verify Mean Value Theorem,if \(f(x)=x^2-4x-3\) in the interval \([a,b]\),where \(a=1\) and \(b=4\)
Verify Mean Value Theorem,if \(f(x)=x^2-4x-3\) in the interval \([a,b]\),where \(a=1\) and \(b=4\)
Updated On: Sep 13, 2023
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Solution and Explanation
The given function is \(f(x)=x^2-4x-3\)
\(f\), being a polynomial function, is continuous in \([1,4]\) and is differentiable in \((1,4)\) whose derivative is \(2x−4\).
\(f(1)=1^2-4\times1-3=-6\)
\(f(4)=(4)^2-4\times4-3=-3\)
\(∴\frac{f(b)-f(a)}{b-a}=\frac{f(4)-f(1)}{4-1}\)
\(=\frac{-3-(-6)}{3}=\frac{3}{3}=1\)
Mean Value Theorem states that there is a point \(c∈(1,4)\) such that \(f'(c)=1\)
\(f'(c)=1\)
\(⇒2c-4=1\)
\(⇒c=\frac{5}{2}\),where \(c=\frac{5}{2}∈(1,4)\)
Hence, Mean Value Theorem is verified for the given function.
\(f\), being a polynomial function, is continuous in \([1,4]\) and is differentiable in \((1,4)\) whose derivative is \(2x−4\).
\(f(1)=1^2-4\times1-3=-6\)
\(f(4)=(4)^2-4\times4-3=-3\)
\(∴\frac{f(b)-f(a)}{b-a}=\frac{f(4)-f(1)}{4-1}\)
\(=\frac{-3-(-6)}{3}=\frac{3}{3}=1\)
Mean Value Theorem states that there is a point \(c∈(1,4)\) such that \(f'(c)=1\)
\(f'(c)=1\)
\(⇒2c-4=1\)
\(⇒c=\frac{5}{2}\),where \(c=\frac{5}{2}∈(1,4)\)
Hence, Mean Value Theorem is verified for the given function.
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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)