Answer to Question #314457 in Calculus for sam

Question

Determine whether or not the following are continuous functions.

1. ƒ(𝑥)=5𝑥+3 2. ƒ(𝑥)=−4𝑥+2

3. ƒ(𝑥)=2𝑥2 +𝑥−3

4. ƒ(𝑥)={2𝑥−3 iƒ 𝑥≥2 −2𝑥 + 2 iƒ 𝑥 < 2

5. ƒ(𝑥)={|𝑥+2|iƒ𝑥G−2 4 iƒ 𝑥 = −2

Answer 1

1. $ƒ(𝑥)=5𝑥+3$

Consider the difference

$f(x+\Delta x)-f(x)=5(x+\Delta x)+3 - (5x+3)=5\Delta x\to 0$ as $\Delta x\to0$, for any $x\in\mathbb{R}$. Therefore, this function is continuous for any $x\in\mathbb{R}$.

2. $ƒ(𝑥)=−4𝑥+2$

Consider the difference

$f(x+\Delta x)-f(x)=-4(x+\Delta x)+2 - (−4𝑥+2)=-4\Delta x\to 0$ as $\Delta x\to0$, for any $x\in\mathbb{R}$. Therefore, this function is continuous for any $x\in\mathbb{R}$.

3. $ƒ(𝑥)=2𝑥^2 +𝑥−3$

Consider the difference

$f(x+\Delta x)-f(x)=2(x+\Delta x)^2 +(x+\Delta x)−3 - (2𝑥^2 +𝑥−3)$

$=4x\Delta x+2(\Delta x)^2\to 0$ as $\Delta x\to0$, for any $x\in\mathbb{R}$.

Therefore, this function is continuous for any $x\in\mathbb{R}$.

4. $ƒ(𝑥)=\left\{ \begin{matrix} 2𝑥−3 & \rm{if}\, 𝑥≥2 \\ −2𝑥 + 2 & \rm{if}\, 𝑥 < 2 \end{matrix} \right.$

For any $x<2$ consider the difference

$f(x+\Delta x)-f(x)=-2(x+\Delta x)+2 - (−2𝑥 + 2)=-2\Delta x\to 0$

as $\Delta x\to0$. Therefore, this function is continuous for any $x<2$.

For any $x>2$ consider the difference

$f(x+\Delta x)-f(x)=2(x+\Delta x)+2 - (2𝑥 + 2)=2\Delta x\to 0$

as $\Delta x\to0$. Therefore, this function is continuous for any $x>2$.

For $x=2$ consider one-sided limits:

$\lim\limits_{\Delta\to+0}f(2+\Delta x)=\lim\limits_{\Delta\to+0}(2(2+\Delta x)-3)=1$

$\lim\limits_{\Delta\to-0}f(2+\Delta x)=\lim\limits_{\Delta\to+0}(-2(2+\Delta x)+2)=-2$

The one-sided limits are not equal, therefore, the function is not continuous at $x=2$.

5. $ƒ(𝑥)=\left\{ \begin{matrix} |𝑥+2| & \rm{if}\, 𝑥\ne -2 \\ 4 & \rm{if}\, 𝑥 = -2 \end{matrix}\right.$

For any $x>-2$ consider the limit

$\lim\limits_{\Delta\to 0}f(x+\Delta x)=\lim\limits_{\Delta\to0}|x+\Delta x+2|$

$=\lim\limits_{\Delta\to0}(x+\Delta x+2)=x+2=f(x)$

Therefore, the function is continuous for any $x>-2$.

For any $x<-2$ consider the limit

$\lim\limits_{\Delta\to 0}f(x+\Delta x)=\lim\limits_{\Delta\to0}|x+\Delta x+2|$

$=\lim\limits_{\Delta\to0}(-x-\Delta x-2)=-x-2=f(x)$

Therefore, the function is continuous for any $x<-2$.

For $x=-2$ consider one-sided limits:

$\lim\limits_{\Delta\to+0}f(-2+\Delta x)=\lim\limits_{\Delta\to+0}|-2+\Delta x+2|=0$

$\lim\limits_{\Delta\to-0}f(-2+\Delta x)=\lim\limits_{\Delta\to-0}|-2+\Delta x+2|=0$

They are not equal to $f(-2)=4$. Therefore, the function is not continuous at $x=-2$.