If the L. M. V. T. holds for the function $f(x)=x+\frac{1}{x}, x \in[1,3]$, then c=

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Aicharya · Accepted Answer

Given $f(x)=x+\frac{1}{x}$ and LMVT holds $f^{\prime}(x)=1-\frac{1}{x^{2}} \Rightarrow f^{\prime}(c)=1-\frac{1}{c^{2}}$ $f(1)=1+1=2 	ext { and } f(3)=3+\frac{1}{3}=\frac{10}{3}$ $	herefore \quad f^{\prime}(c)=1-\frac{1}{c^{2}}=\frac{\frac{10}{3}-2}{(3-1)} \Rightarrow 1-\frac{1}{c^{2}}=\frac{4}{3(2)}=\frac{2}{3}$ $	herefore \quad \frac{1}{c^{2}}=1-\frac{2}{3}=\frac{1}{3} \Rightarrow c^{2}=3 \Rightarrow c=\pm \sqrt{3}$

If the L. M. V. T. holds for the function $f(x)=x+\frac{1}{x}, x \in[1,3]$, then c=

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