Verify mean value theorem f (x) =x3, x€(3, 4)
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Answer:
The given function is f(x)=x
3
−5x
2
−3x
f being a polynomial function, so it is continuous in [1,3] and is differentiable in (1,3) whose derivative is 3x
2
−10x−3.
f(1)=1
3
−5⋅1
2
−3⋅1=−7,f(3)=3
3
−5⋅3
2
−3⋅3=−27
∴
b−a
f(b)−f(a)
=
3−1
f(3)−f(1)
=
3−1
−27−(−7)
=−10
Mean Value Theorem states that there exist a point c∈(1,3) such that f
′
(c)=−10
⇒3c
2
−10c−3=−10
⇒3c
2
−10c+7=0
⇒3c
2
−3c−7c+7=0
⇒3c(c−1)−7(c−1)=0
⇒(c−1)(3c−7)=0
⇒c=1,
3
7
, where c=
3
7
∈(1,3)
Hence, Mean Value Theorem is verified for the given function and c=
3
7
∈(1,3) is the point for which f
′
(c)=0
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