Question

Find the point of discontinuity of $y=f\left(u\right)$, where $f\left(u\right)=\frac{3}{2u^2+5u-3}$ and $u=\frac{1}{x+2}$

Answer 1

$u=\frac{1}{x+2}$ for $u$ to be real value

$\begin{array}{c}x+2\ne 0\\ or,x\ne -2\end{array}$

$\begin{array}{c}f\left(u\right)=\frac{3}{2u^2+5u-3}=\frac{3}{2\frac{1}{{(x+2)}^2}+\frac{5}{x+2}-3}\\ =\frac{3{(x+2)}^2}{2+5(x+2)-3{(x+2)}^2}\end{array}$

reducing denominator:

$\begin{array}{c}-3\left[x^2+4x+4\right]+5x+10+2\\ =-3x^2-12x-12+5x+10+2\\ =-3x^2-7x\\ \therefore f\left(x\right)=\frac{3{(x+2)}^2}{-3x^2-7x}=\frac{3{(x+2)}^2}{-x\left[3x+7\right]}\end{array}$

for $f\left(u\right)$ to be defined

$\begin{array}{c}x\ne 0\\ 3x+7\ne 0\\ x\ne \frac{-7}{3}\\ \therefore at,x=-2,0,\frac{-7}{3}\end{array}$

funtion is discontinuous as it is undefined at those points.