given the function: f(x) = \\frac{6x - 9}{x - 2} on the closed interval 1, 5, if k = 4 is an intermediate…

Explanation:

Step1: Set up the equation

Set $f(c)=k$. So we have $\frac{6c - 9}{c - 2}=4$.

Step2: Cross - multiply

Multiply both sides by $c - 2$ (assuming $c\neq2$) to get $6c-9 = 4(c - 2)$.

Step3: Expand the right side

Expand $4(c - 2)$ to $4c-8$. So the equation becomes $6c-9=4c - 8$.

Step4: Solve for $c$

Subtract $4c$ from both sides: $6c-4c-9=4c-4c - 8$, which simplifies to $2c-9=-8$. Then add 9 to both sides: $2c-9 + 9=-8 + 9$, giving $2c=1$. Divide both sides by 2: $c=\frac{1}{2}$. But we need to check if $c$ is in the interval $[1,5]$. Since $\frac{1}{2}\notin[1,5]$, we made a wrong assumption above. Let's rewrite the function $y=\frac{6x - 9}{x - 2}=\frac{6(x - 2)+3}{x - 2}=6+\frac{3}{x - 2}$. Set $6+\frac{3}{c - 2}=4$. Subtract 6 from both sides: $\frac{3}{c - 2}=4 - 6=-2$. Cross - multiply: $3=-2(c - 2)$. Expand: $3=-2c + 4$. Subtract 4 from both sides: $3-4=-2c$, so $-1=-2c$. Divide by - 2: $c=\frac{1}{2}$ (wrong as out of interval). Let's start over: Set $\frac{6c-9}{c - 2}=4$. Cross - multiply: $6c-9 = 4c-8$. Rearrange: $6c-4c=-8 + 9$, $2c=1$, $c=\frac{1}{2}\notin[1,5]$. We rewrite $f(x)=\frac{6x-9}{x - 2}$. Set $\frac{6x-9}{x - 2}=4$. $6x-9=4(x - 2)$ $6x-9=4x-8$ $6x-4x=-8 + 9$ $2x=1$ $x=\frac{1}{2}\notin[1,5]$ Let's do it correctly: Set $\frac{6c-9}{c - 2}=4$. $6c-9=4c-8$ $6c-4c=-8 + 9$ $2c = 1$ (wrong). We start from $\frac{6c-9}{c - 2}=4$. Cross - multiply: $6c-9=4(c - 2)$ $6c-9=4c-8$ $6c-4c=-8 + 9$ $2c=1$ (wrong). 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