Answer to Question #310474 in Calculus for Mom

Question #310474

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)f(x) is continuous on interval [a,b][a, b] , and differentiable on (a,b)(a, b) .


Then there exists cc such that a<c<ba < c <b and 


f(c)=f(b)f(a)baf'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}


In the question the function is missing, so let us choose, the function as


f(x)=x3+2x3f\left( x \right) = {x^3} + 2x - 3 which is continuous with in the interval [0,8][0,8] and differentiable within (0,8)(0, 8) ,

then for 0<c<80 < c <8, we have


f(c)=f(8)f(0)80f'\left( c \right) = \frac{{f\left( 8 \right) - f\left( 0 \right)}}{{8 - 0}}


f(c)=[(0)32(0)3][(8)32(8)3]80f'\left( c \right) = \frac{{\left[ {{{\left( 0 \right)}^3} - 2\left( 0 \right) - 3} \right] - \left[ {{{\left( 8 \right)}^3} - 2\left( 8 \right) - 3} \right]}}{{8 - 0}}\\


f(c)=34938f'\left( c \right) = \frac{{ - 3 - 493}}{8}\\


f(c)=62f'\left( c \right) = 62


Now


f(x)=x3+2x3f\left( x \right) = {x^3} + 2x - 3\\


f(x)=3x2+2f'\left( x \right) = 3{x^2} + 2\\


f(c)=3c2+2f'\left( c \right) = 3{c^2} + 2


Therefore,


3c2+2=623{c^2} + 2=62


3c2=6223{c^2} =62-2


c2=603=20c^2=\frac{60}{3}=20


c=2\sqrt 5 \ (ignoring negative value since we need to focus within the interval [0,8][0, 8] )


since we can see that


0<25<80<2\sqrt 5 < 8


Hence required value of cc is 252\sqrt 5




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