Question

Find the relationship between a and b, so that the function $f(x)$ is defined by $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {ax + 2,{\text{ }}if{\text{ }}x \leqslant 4} \\
  {bx + 4,{\text{ }}if{\text{ }}x > 4}
\end{array}} \right.$ is continuous at $x = 3$.

Answer 1

Hint: We have to apply limits according to the given conditions and solve the problem step by step to find the relationship between \[a\] and \[b\].
So, we have to use the first function as left hand limit and similarly the second function as right hand limit. Finally equate both limits.
At last, we get the required answer.

Complete step-by-step solution:
The limit for a function $f(x)$ can be written as \[\mathop {\lim }\limits_{x \to a} f(x)\]
It is given that the question stated as the function is continuous at $x = 3$
Now we have to find out the left hand limit
That is the left hand limit found for a function $x = a$ by substituting the value of $x - h$ instead of the value of \[x\].
To find the left hand limits when the function $f(x)$ is continuous at $x = 3$.
So we can write it as, $\mathop {\lim }\limits_{x \to 3 - } f\left( x \right) = \mathop {\lim }\limits_{x \to 3 - } \left( {ax + 2} \right)$
On substituting $x = 3 - h$ and we get
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} a(3 - h) + 2$
On multiplying the bracket term and we get
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} 3a - ah + 2$
By applying the limit $h = 0$,
$ \Rightarrow 3a - a(0) + 2$
Hence we get the left hand limit
$ \Rightarrow 3a + 2$
Also, we have to find the right hand limit,
Here, right hand limit is found for a function $x = a$ by substituting the value of $x + h$ instead of the value of \[x\]
To find the right hand limits when the function $f(x)$ is continuous at$x = 3$.
So we can write it as, $\mathop {\lim }\limits_{x \to 3 - } f\left( x \right) = \mathop {\lim }\limits_{x \to 3 + } (bx + 4)$
On substituting $x = 3 + h$ and we get
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} b(3 + h) + 4$
On multiplying the bracket terms and we get
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} 3b + bh + 4$
By applying the limit $h = 0$,
$ \Rightarrow 3b + b(0) + 4$
Hence, we get the right hand limit as,
$ \Rightarrow 3b + 4$
Now, we have to equating the left hand limits and right hand limits of the functions,
LHS = RHS = $f(x)$
So we can write it as,
$3a + 2 = 3b + 4$
Taking the integer as RHS and variable terms as LHS we get,
$3a - 3b = 4 - 2$
Taking the common term as LHS and subtract the RHS we get,
$ \Rightarrow 3(a - b) = 2$

Hence, the relationship between \[a\] and \[b\] is determined as, $a - b = \dfrac{2}{3}$ or $a = b + \dfrac{2}{3}$.

Note: If the function is continuous, then it is continuous of all points.
If a point has a function value that is not equal to its limit, because it \[x\] approaches that point is called the point of discontinuity.
Otherwise, a function that has points of discontinuity is known as discontinuous.