Mean Value Theorem for Continuous Functions
Key Questions
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The value of
#c# is#sqrt{3}# .Let us look at some details.
M.V.Thm. states that there exists
#c# in (0,3) such that#f'(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 function#f# is continuous on#[a,b]# and differentiable on#(a,b)# ,
then there exists c in#(a,b)# such that#f'(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