Question

Find 'c' of Lagrange's mean-value theorem for
(i) $f\left(x\right)=(x^3-3x^2+2x)$ on $[0,\frac{1}{2}]$
(ii) $f\left(x\right)=\sqrt{25-x^2}$ on $[0,5]$
(iii) $f\left(x\right)=\sqrt{x+2}$ on $[4,6]$

Answer 1

$\left(i\right)f\left(x\right)=x^3-3x^2+2x$ on $[0,\frac{1}{2}]$

Since $f\left(x\right)$ is continuous on [$0,\frac{1}{2}$] and differentials on $[0,\frac{1}{2}]$.

Mean value theorm can be applied.

$f^{\prime }\left(x\right)=x^3-3x^2+2x$

$f\left(0\right)=0$

$f\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^3-{\left(\frac{1}{2}\right)}^2+2\left(\frac{1}{2}\right)=\frac{1}{8}-\frac{3}{4}+1=\frac{8-6+1}{8}=\frac{3}{8}$

According to MVT,

$\frac{\left[f\right(b)-f(a\left)\right]}{(b-a)}=f^{{}^{\prime }}\left(c\right)\Rightarrow \frac{\left(\frac{3}{8}\right)-0}{(\frac{1}{2}-0)}=3c^2-6c+2\Rightarrow (3c^2-6c+2)=\frac{3\times 2}{4}=\frac{3}{4}\Rightarrow {12c}^2-24c+8=3\Rightarrow {12c}^2-24c+5=0\Rightarrow {12c}^2-24c+5=0\Rightarrow c=\frac{24\pm \sqrt{{\left(24\right)}^2-4\left(12\right)\left(5\right)}}{24}=\frac{24\pm \sqrt{576-240}}{24}=\frac{24\pm \sqrt{336}}{24}\Rightarrow c=\frac{1\pm \sqrt{21}}{6}=0.236or1.764$

Since the domain of $f\left(x\right)$ given $[0,\frac{1}{2}],C$ must be $0.236$

$\left(ii\right)$

$f\left(x\right)=\sqrt{25-x^2}$ on $[0,5]$

$f^{{}^{\prime }}\left(x\right)=\frac{1}{2\sqrt{25-x^2}}(-2x)=-\frac{x}{\sqrt{25-x^2}}\therefore \frac{\left[f\right(b)-f(a\left)\right]}{(b-a)}=f^{{}^{\prime }}\left(c\right)\Rightarrow \frac{\left[f\right(5)-f(0\left)\right]}{(5-0)}=\frac{-c}{\sqrt{25-c^2}}\Rightarrow \frac{\left(\sqrt{25-25}\right)-\sqrt{25-0}}{5}=\frac{-c}{\sqrt{25-c^2}}\Rightarrow \frac{0-5}{5}=\frac{-c}{\sqrt{25-c^2}}\Rightarrow -1=\frac{-c}{\sqrt{25-c^2}}\Rightarrow 25.c^2=c^2\Rightarrow 2c^2=2.5\Rightarrow c=\frac{5}{2}$

$\left(iii\right)$

$f\left(x\right)=\sqrt{x+2}$ on $[4,6]$

$f^{\prime }x=\frac{1}{2\sqrt{x+2}}$

$\Rightarrow \frac{\left[f\right(b)-f(a\left)\right]}{(b-a)}=f^{{}^{\prime }}\left(c\right)=\frac{\left[f\right(6)-f(4\left)\right]}{(6-4)}=\frac{1}{2\sqrt{c+2}}\Rightarrow \frac{\left(\sqrt{8}\right)-\sqrt{6}}{5}=\frac{1}{2\sqrt{c+2}}\Rightarrow (\sqrt{8}+\sqrt{6})=\frac{1}{\sqrt{c+2}}\Rightarrow 8+6+8\sqrt{3}=\frac{1}{(c+2)}\Rightarrow 14+8\sqrt{3}=\frac{1}{c+2}\Rightarrow c+2=\frac{1}{14+8\sqrt{3}}\times \frac{14.8\sqrt{3}}{14.8\sqrt{3}}=\frac{14-8\sqrt{3}}{196-192}=\frac{14-8\sqrt{3}}{4}\Rightarrow c+2=\frac{7-4\sqrt{3}}{2}\Rightarrow c=\frac{7}{2}-2-2\sqrt{3}=\frac{3}{2}-2\sqrt{3}$