Respuesta :
Answer:
The answer is "[tex]{c=-6+3\sqrt{14}[/tex]"
Step-by-step explanation:
Given:
[tex]\to F(x) = \frac{x}{x+6} , \ [1, 12][/tex]
[tex]\to f'(c)=\frac{f(b)-f(a)}{b-a}[/tex]
As per the given intervals:
[tex][1,12]\\\\ a=1 \\ b=12.[/tex]
Calculating the values of f(a) and f(b).
[tex]\to f(1)=\frac{1}{1+6}= \frac{1}{7} \\\\ \to f(2)=\frac{12}{12+6}= \frac{12}{18}=\frac{2}{3} \\\\ \to f(x)=\frac{x}{x+6}\\\\[/tex]
[tex]\to f\:'\left(x\right)=\frac{d}{dx}\left(\frac{x}{x+6}\right)\: \\\\[/tex]
[tex]=\frac{\frac{d}{dx}\left(x\right)\left(x+6\right)-\frac{d}{dx}\left(x+6\right)x}{\left(x+6\right)^2} \\\\=\frac{1\cdot \left(x+6\right)-1\cdot \:x}{\left(x+6\right)^2}\\\\[/tex]
[tex]\to f'(x)=\frac{6}{\left(x+6\right)^2}[/tex]
Calculating the value of [tex]\ f'(c):[/tex]
[tex]\to \mathbf{f\:'\left(c\right)=\frac{6}{\left(c+6\right)^2}}[/tex]
Replacement of the values in the mean theorem of value now.
[tex]\to \frac{6}{\left(c+6\right)^2}=\frac{\frac{2}{3}-\frac{1}{7}}{12-1}\\\\\to \frac{6}{\left(c+6\right)^2}=\frac{\frac{11}{21}}{11}\\\\\to \frac{6}{\left(c+6\right)^2}=\frac{1}{21}\\\\[/tex]
Apply the cross multiplication:
[tex]\to (c+6)^2=126 \\\\\to c^2+12c +36= 126\\\\\to c^2+12c +36-126= 126-126 \\\\\to c^2+12c -90= 0\\\\\to c=\frac{-12\pm \sqrt{12^2-4\cdot \:1\left(-90\right)}}{2\cdot \:1}\\\\ \to \mathbf{c=-6+3\sqrt{14}} \\\\[/tex]
