Given
[tex]f\mleft(x\mright)=x^3-2x²+x+1,\left[-2,2\right][/tex]
Find
value of c
Explanation
mean value theorem states that " Let f be continuous over the closed interval [a , b] and differentiable over the open interval (a , b). Then , there exist atleast one poin c which belongs to (a , b) such that
[tex]f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}[/tex]
so ,
[tex]\begin{gathered} f(x)=x^3-2x^2+x+1 \\ f^{\prime}(x)=3x^2-4x+1 \\ f^{\prime}(c)=3c^2-4c+1e^{\placeholder{⬚}} \end{gathered}[/tex]
and
[tex]\begin{gathered} f(b)=f(2)=2^3-2(2)^2+2+1=8-8+3=3 \\ f(a)=f(-2)=(-2)^3-2(-2)^2-2+1=-8-8-1=-17 \end{gathered}[/tex]
so ,
[tex]\begin{gathered} 3c^2-4c+1=\frac{3-(-17)}{2-(-2)} \\ \\ 3c^2-4c+1=\frac{20}{4} \\ \\ 3c^2-4c+1=5 \\ 3c^2-4c-4=0 \\ 3c^2-6c+2c-4=0 \\ 3c(c-2)+2(c-2)=0 \\ (3c+2)(c-2)=0 \\ c=-\frac{2}{3},2 \end{gathered}[/tex]
Final Answer
Therefore , the values of c are -2/3 and 2
