Mean Value Theorem for Continuous Functions
Key Questions
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The value of
cc issqrt{3}√3 .Let us look at some details.
M.V.Thm. states that there exists
cc in (0,3) such thatf'(c)={f(3)-f(0)}/{3-0} .Let us find such
c .The left-hand side is
f'(c)=3c^2+1 .The right-hand side is
{f(3)-f(0)}/{3-0}={29-(-1)}/{3}=10 .By setting them equal to each other,
3c^2+1=10 Rightarrow 3x^2=9 Rightarrow x^2=3 Rightarrow x=pm sqrt{3} Since
0<c<3 ,c=sqrt{3} .I hope that this was helpful.
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Actually, Rolle's Theorem require differentiablity, and it is a special case of Mean Value Theorem.
Please watch this video for more details.
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Mean Value Theorem
If a functionf is continuous on[a,b] and differentiable on(a,b) ,
then there exists c in(a,b) such thatf'(c)={f(b)-f(a)}/{b-a} .
Questions
Graphing with the First Derivative
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Interpreting the Sign of the First Derivative (Increasing and Decreasing Functions)
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Identifying Stationary Points (Critical Points) for a Function
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Identifying Turning Points (Local Extrema) for a Function
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Classifying Critical Points and Extreme Values for a Function
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Mean Value Theorem for Continuous Functions