Chapter 2.3, Problem 50E

To determine

To find: The value of $a$ is such that the function $f\left(x\right)=\{\begin{array}{l}{x}^2+x+a,x<1\\ x^3,x\ge 1\end{array}$ is continuous.

Answer to Problem 50E

The value of $a$ is $-1$ such that the function $f\left(x\right)=\{\begin{array}{l}{x}^2+x+a,x<1\\ x^3,x\ge 1\end{array}$ is continuous.

Explanation of Solution

Given information: The function $f\left(x\right)=\{\begin{array}{l}{x}^2+x+a,x<1\\ x^3,x\ge 1\end{array}$ is continuous.

Calculation:

The function $f\left(x\right)=\{\begin{array}{l}{x}^2+x+a,x<1\\ x^3,x\ge 1\end{array}$ is continuous. So, the function is continuous at all the points of $x$ . The function is also continuous at $x=1$ . So, $\underset{x\to 1}{\lim }f\left(x\right)=f\left(1\right)$.

The function $f\left(x\right)=x^2+x+a$ corresponds to all the values of $x<1$.

Calculate the left hand limit of function at $x=1$.

  $\begin{array}{c}\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{-}}{\lim }\left(x^2+x+a\right)\\ =\underset{x\to 1^{-}}{\lim }\left(x^2\right)+\underset{x\to 1^{-}}{\lim }\left(x\right)+\underset{x\to 1^{-}}{\lim }\left(a\right)\\ ={\left(1\right)}^2+1+a\\ =2+a\end{array}$

The function $f\left(x\right)=x^3$ corresponds to all the values of $x\ge 1$.

Calculate the right hand limit of function at $x=1$.

  $\begin{array}{c}\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }\left(x^3\right)\\ =1^3\\ =1\end{array}$

It is given that the function is continuous at $x=1$ . So, left hand limit will be equal to right hand limit of the function as:

  $\begin{array}{c}\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)\\ 2+a=1\\ a=-1\end{array}$

Therefore, the value of $a$ is $-1$ such that the function $f\left(x\right)=\{\begin{array}{l}{x}^2+x+a,x<1\\ x^3,x\ge 1\end{array}$ is continuous.