We need to satisfy the following conditions:
1. The function is continuous on the closed interval [a,b]
2. f is differentiable on the open interval (a,b).
Let's check every condition:
[tex]\begin{gathered} f(x)=4x^3 \\ \lbrack1,2\rbrack \end{gathered}[/tex]
The function is continuous since there are no any restrictions.
[tex]f^{\prime}(x)=12x^2[/tex]
The derivative of the function exists over the interval (1,2)
Therefore:
Answer:
Yes, The mean value theorem can be applied
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[tex]\begin{gathered} f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\ f(b)=f(2)=4(2)^3=32 \\ f(a)=f(1)=4(1)^3=4 \\ so\colon \\ \frac{32-4}{2-1}=28 \end{gathered}[/tex][tex]\begin{gathered} f^{\prime}(c)=28 \\ 28=12x^2 \\ x^2=\frac{28}{12} \\ x=\sqrt[]{\frac{7}{3}} \\ \end{gathered}[/tex]
Therefore:
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