Question

F(x) = x^3 - 2x² + x + 1 on [-2,2]

Answer 1

Given

`f\mleft(x\mright)=x^3-2x²+x+1,\left[-2,2\right]`

Find

value of c

Step-by-step 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

`f^(\prime)(c)=(f(b)-f(a))/(b-a)`

so ,

`\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}`

and

`\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}`

so ,

`\begin{gathered} 3c^2-4c+1=(3-(-17))/(2-(-2)) \\ \\ 3c^2-4c+1=(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=-(2)/(3),2 \end{gathered}`

Final Answer

Therefore , the values of c are -2/3 and 2