Estimate the values of c that satisfy the conclusion of the Mean Value Theorem on the interval (0,8). Enter your answer as a comma-separated list. Round to one decimal.
1
Expert's answer
2022-03-15T15:13:24-0400
Solution
According to the mean value theorem,
If f(x) is continuous on interval [a,b] , and differentiable on (a,b) .
Then there exists c such that a<c<b and
f′(c)=b−af(b)−f(a)
In the question the function is missing, so let us choose, the function as
f(x)=x3+2x−3 which is continuous with in the interval [0,8] and differentiable within (0,8) ,
then for 0<c<8, we have
f′(c)=8−0f(8)−f(0)
f′(c)=8−0[(0)3−2(0)−3]−[(8)3−2(8)−3]
f′(c)=8−3−493
f′(c)=62
Now
f(x)=x3+2x−3
f′(x)=3x2+2
f′(c)=3c2+2
Therefore,
3c2+2=62
3c2=62−2
c2=360=20
c=2\sqrt 5 \ (ignoring negative value since we need to focus within the interval [0,8] )
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