Question

(Answer: c=+-1)
Q.8. In the mean value theorem f (b) - f (a) = (b - a) f ' (c), (a < c < b); if f (x) = $x^3-3x-1anda=-\frac{11}{7},b=\frac{13}{7}$, find the value of c .

Answer 1

Dear student
$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that},\mathrm{by}\mathrm{Lagranges}\mathrm{mean}\mathrm{value}\mathrm{theorem} \\ \mathrm{f}\text{'}\left(\mathrm{c}\right)=\frac{\mathrm{f}\left(\mathrm{b}\right)-\mathrm{f}\left(\mathrm{a}\right)}{\mathrm{b}-\mathrm{a}} \\ \mathrm{here}\mathrm{a}=\frac{-11}{7}\mathrm{and}\mathrm{b}=\frac{13}{7} \\ \mathrm{Given}:\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3\mathrm{x}-1 \\ \mathrm{So},\mathrm{f}\text{'}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-3 \\ \mathrm{Since}\mathrm{a}\mathrm{polynomial}\mathrm{function}\mathrm{is}\mathrm{everywhere}\mathrm{continuous}\mathrm{and}\mathrm{differentiable} \\ \mathrm{So},\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{continuous}\mathrm{on}\left[\frac{-11}{7},\frac{13}{7}\right]\mathrm{and}\mathrm{diffferentiable}\mathrm{on}\left(\frac{-11}{7},\frac{13}{7}\right) \\ \mathrm{Thus},\mathrm{both}\mathrm{the}\mathrm{conditions}\mathrm{of}\mathrm{LMV}\mathrm{theorem}\mathrm{are}\mathrm{satisfied}. \\ \mathrm{So},\mathrm{there}\mathrm{exists}\mathrm{atleast}\mathrm{one}\mathrm{real}\mathrm{number}\mathrm{c}\in \left(\frac{-11}{7},\frac{13}{7}\right)\mathrm{such}\mathrm{that} \\ \mathrm{f}\text{'}\left(\mathrm{c}\right)=\frac{\mathrm{f}\left(\frac{13}{7}\right)-\mathrm{f}\left({\displaystyle \frac{-11}{7}}\right)}{\frac{13}{7}-\frac{(-11)}{7}} \\ \mathrm{Now},\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3\mathrm{x}-1 \\ \mathrm{So},\mathrm{f}\text{'}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-3,\mathrm{f}\left(\frac{-11}{7}\right)={\left(\frac{-11}{7}\right)}^3-3\left(\frac{-11}{7}\right)-1=\frac{-1331}{343}+\frac{33}{7}-1=\frac{-57}{343} \\ \mathrm{and}\mathrm{f}\left(\frac{13}{7}\right)={\left(\frac{13}{7}\right)}^3-3\left(\frac{13}{7}\right)-1=\frac{2197}{343}-\frac{39}{7}-1=\frac{-57}{343} \\ \mathrm{So},\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\frac{-57}{343}-\left(\frac{-57}{343}\right)}{{\displaystyle \frac{24}{7}}}=0 \\ \Rightarrow 3{\mathrm{x}}^2-3=0 \\ \Rightarrow 3{\mathrm{x}}^2=3 \\ \Rightarrow {\mathrm{x}}^2=1 \\ \Rightarrow \boxed{\mathbf{x}\mathbf{=}\mathbf{\pm }\mathbf{1}} \end{gathered}$$
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