Question
Examine the continuity of the function
$\text{f}\text{(x)}=\begin{cases}3\text{x}-2 &, \text{ if x} \leq 0\\\text{x}+1 &, \text{ if x} > 0\end{cases}\text{at x}=0$
Also sketch the graph of this function.

Answer

Given function is, $\text{f}\text{(x)}=\begin{cases}3\text{x}-2 &, \text{ if x} \leq 0\\\text{x}+1 &, \text{ if x} > 0\end{cases}\text{at x}=0\ ...(\text{ii})$ We need to check whether f(x) is continuous at x = 0 or not. For this we need to check L.H.L, R.H.L and value of function at x = 0 Clearly, f(0) = 3*0 - 2 = -2 [from equation ii] $\text{L.H.L}=​​\lim\limits_{\text{h} \rightarrow 0}(0-\text{h})=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h})$ $=\lim\limits_{\text{h} \rightarrow 0}\big\{3(-\text{h})-2\big\}=-2$ $\text{R.H.L}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h})=\lim\limits_{\text{h} \rightarrow 0}\text{f(h)}$ $=\lim\limits_{\text{h} \rightarrow 0}\big\{\text{h}+1\big\}=0+1=1$ As, $\text{L.H.L}\neq\text{R.H.L}$ $\therefore\ \text{f(x)}$ is discontinuous at x = 0 This can also be proved by plotting f(x) on cartesian plane. For x >0 ,we need to plot y = x + 1 put y = 0, we get x = -1 and for second point we put x = 0 and thus get y = 1 Two points are enough to plot the straight line. Two coordinates are (-1, 0) and (0, 1) For $\text{x}\leq0,$ we need to plot y = 3x - 2 put x = 0 then y = -2 On putting y = 0 we get $\text{x}=\frac{2}{3}$ Two coordinates are (0, -2) and $\Big(\frac{2}{3},0\Big)$ It can be seen from graph that there is breakage in curve at (0, 0) Thus, it is discontinuous at x = 0

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

By examining the chest X ray, the probability that TB is detected when a person is actually suffering is $0.99$. The probability of an healthy person diagnosed to have TB is $0.001$. In a certain city, $1$ in $1000$ people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?
Solve the following differential equations:$\tan\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{x}+\text{y})+\sin(\text{x}-\text{y})$
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{24\text{x}^3}{(1+\text{x}^2)^4}\text{ dx}$
Evaluate the following integrals:$\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{8-\sin2\text{x}}}\text{ dx}$
Find the equation of the passing throught the point (1, 2, 1) and perpendicular to the joining the point (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane.
Find the vector equation of the plane passing through the points (1, 1, 1), (1, -1, 1) and (-7, -3, -5)
Find the intervals in which the following functions are increasing or decreasing.
$\text{f}(\text{x})=\frac{3}{10}\text{x}^4-\frac{4}{5}\text{x}^3-3\text{x}^2+\frac{36}{5}\text{x}+11$
The relation S defined on the set R of all real number by the rule aSb iff a ≥ b is:
  1. An equivalence relation.
  2. Reflexive, transitive but not symmetric.
  3. Symmetric, transitive but not reflexive.
  4. Neither transitive nor reflexive but symmetric.
Evaluate:
$\int\frac{1}{\sin^{4}\text{x} +\sin^{2}\text{x}\cos^{2}\text{x}+\cos^{4}\text{x}}\text{dx}$
Evaluate the following integrals:
$\int\frac{\text{x}}{\text{x}^4+2\text{x}^2+3}\text{dx}$