Find the value of c guaranteed by the Intermediate Value Theorem. $$f ( x ) = \frac { x ^ { 2 } - 5 x } { x - 3 } , \left[ \frac { 9 } { 2 } , 18 \right] , f ( c ) = 6$$
A) 11
B) 2
C) 1
D) 9
E) 10

Question

Quizplus · Accepted Answer

First, we note that $f(x)$ is not defined at $x=3$, but since $3$ is not in the given interval, we can ignore it. 
Since $f(x)$ is continuous on $[\frac{9}{2},18]$, by the Intermediate Value Theorem, there exists a value $c$ in $[\frac{9}{2},18]$ such that $f(c) = 6$ if and only if $f(x)$ takes on all values between $f(\frac{9}{2})$ and $f(18)$ on this interval.
We calculate $f(\frac{9}{2})=-10\frac{1}{2}$ and $f(18)=39$.
Since $6$ is between $-10\frac{1}{2}$ and $39$, it follows that there exists $c$ in $[\frac{9}{2},18]$ such that $f(c) = 6$. 
We could find the value of $c$ by solving $f(c)=6$ algebraically, but this is not necessary to answer the question.

Find the Value of C Guaranteed by the Intermediate Value f(x)=x2−5xx−3,[92,18],f(c)=6f ( x ) = \frac { x ^ { 2 } - 5 x } { x - 3 } , \left[ \frac { 9 } { 2 } , 18 \right] , f ( c ) = 6f(x)=x−3x2−5x​,[29​,18],f(c)=6

Find the Value of C Guaranteed by the Intermediate Value $f ( x ) = \frac { x ^ { 2 } - 5 x } { x - 3 } , \left[ \frac { 9 } { 2 } , 18 \right] , f ( c ) = 6$