Monotonicity, Maxima and Minima
10.0 Maxima and Minima of Discontinuous functions
10.0 Maxima and Minima of Discontinuous functions
Let us assume a discontinuous function $f(x)$ which is discontinuous at $x=a$. In order to find the maxima or minima of anydiscontinuous function, we have to check at the point of discontinuity only using the following conditions:
- At $x=a$, if the function satisfies the condition $$f(a) \geqslant f(a + h){\text{ and }}f(a) \geqslant f(a - h)$$ then $x=a$ is the point of maxima. In the figure, $f(a-h)$ represents the value of the function just below $x=a$ and $f(a+h)$ represents the value of the function just above $x=a$.
- At $x=a$, if the function satisfies the condition $$f(a) \leqslant f(a + h){\text{ and }}f(a) \leqslant f(a - h)$$ then $x=a$is the point of minima. In the figure, $f(a-h)$ represents the value of the function just below $x=a$ and $f(a+h)$ represents the value of the function just above $x=a$.
Question 16. Find the range of $b$ if the function \[f(x) = \left\{ {\begin{array}{c}{2x - 3}&{0 < x \leqslant 2} \\ {3x + b}&{2 < x \leqslant 4}\end{array}} \right.\] has a point of maxima at $x=2$.
Solution: As from the function, it is clear that it is discontinuous at $x=2$ and it is given that $x=2$ is a point of maxima. Therefore, it must satisfy the condition $$f(2) \geqslant f(2 + h){\text{ and }}f(2) \geqslant f(2 - h)$$ From the function, we can calculate that $$\begin{equation} \begin{aligned} f(2) = 2 \times 2 - 3 = 4 - 3 = 1 \\ f(2 + h) = 3(2 + h) + b\;(\because {\text{ for }}2 < x \leqslant 4,{\text{ }}f(x) = 3x + b) \\ f(2 - h) = 2(2 - h) - 3\;(\because {\text{ for }}0 < x \leqslant 2,{\text{ }}f(x) = 2x - 3) \\\end{aligned} \end{equation} $$ Put these values in the condition, we get $$1 \geqslant 3(2 + h) + b\;{\text{and }}1 \geqslant 2(2 - h) - 3$$ As $h \sim 0$, we can write $$\begin{equation} \begin{aligned} 1 \geqslant 6 + b \\ 1 - 6 \geqslant b \\ b \leqslant - 5 \\ \Rightarrow b \in ( - \infty , - 5] \\\end{aligned} \end{equation} $$
