Question

For \\(g(x)=\\frac{x-4}{x-3}, we can use the mean value theorem on [4, 6], Hence determine \\(c\\)

Answer 1

ANSWER: $c=3+\sqrt { 3 }$

EXPLANATION: On the segment [4,6] the function$g(x)=\frac { x-4 }{ x-3 } =1-\frac { 1 }{ x-3 }$ is differentiable, therefore by the mean value theorem , exists a point  $(c,g(c))$ which the tangent line is parallel to the secant connecting the secant connecting the endpoints of the graph. Secant slope is $\frac { g(4)-g(6) }{ 4-6 } =\frac { 0-\frac { 2 }{ 3 } }{ -2 } =\frac { 1 }{ 3 }$ . Tangent slope is

$g'(c)= \frac { 1 }{ { \left( c-3 \right) }^{ 2 } } \ =\frac { 1 }{ 3 }$ . Hence, $c=3+\sqrt { 3 } ,\quad g(c)=1-\frac { 1 }{ \sqrt { 3 } }$