Question #314457

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


1
Expert's answer
2022-03-20T10:12:10-0400

1. ƒ(𝑥)=5𝑥+3ƒ(𝑥)=5𝑥+3

Consider the difference

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


2. ƒ(𝑥)=−4𝑥+2ƒ(𝑥)=−4𝑥+2

Consider the difference

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


3. ƒ(𝑥)=2𝑥2+𝑥−3ƒ(𝑥)=2𝑥^2 +𝑥−3

Consider the difference

f(x+Δx)−f(x)=2(x+Δx)2+(x+Δx)−3−(2𝑥2+𝑥−3)f(x+\Delta x)-f(x)=2(x+\Delta x)^2 +(x+\Delta x)−3 - (2𝑥^2 +𝑥−3)

=4xΔx+2(Δx)2→0=4x\Delta x+2(\Delta x)^2\to 0 as Δx→0\Delta x\to0, for any x∈Rx\in\mathbb{R}.

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


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

For any x<2x<2 consider the difference

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

as Δx→0\Delta x\to0. Therefore, this function is continuous for any x<2x<2.

For any x>2x>2 consider the difference

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

as Δx→0\Delta x\to0. Therefore, this function is continuous for any x>2x>2.

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

lim⁡Δ→+0f(2+Δx)=lim⁡Δ→+0(2(2+Δx)−3)=1\lim\limits_{\Delta\to+0}f(2+\Delta x)=\lim\limits_{\Delta\to+0}(2(2+\Delta x)-3)=1

lim⁡Δ→−0f(2+Δx)=lim⁡Δ→+0(−2(2+Δx)+2)=−2\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=2x=2.


5. ƒ(𝑥)={∣𝑥+2∣if 𝑥≠−24if 𝑥=−2ƒ(𝑥)=\left\{ \begin{matrix} |𝑥+2| & \rm{if}\, 𝑥\ne -2 \\ 4 & \rm{if}\, 𝑥 = -2 \end{matrix}\right.

For any x>−2x>-2 consider the limit

lim⁡Δ→0f(x+Δx)=lim⁡Δ→0∣x+Δx+2∣\lim\limits_{\Delta\to 0}f(x+\Delta x)=\lim\limits_{\Delta\to0}|x+\Delta x+2|

=lim⁡Δ→0(x+Δx+2)=x+2=f(x)=\lim\limits_{\Delta\to0}(x+\Delta x+2)=x+2=f(x)

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

For any x<−2x<-2 consider the limit

lim⁡Δ→0f(x+Δx)=lim⁡Δ→0∣x+Δx+2∣\lim\limits_{\Delta\to 0}f(x+\Delta x)=\lim\limits_{\Delta\to0}|x+\Delta x+2|

=lim⁡Δ→0(−x−Δx−2)=−x−2=f(x)=\lim\limits_{\Delta\to0}(-x-\Delta x-2)=-x-2=f(x)

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

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

lim⁡Δ→+0f(−2+Δx)=lim⁡Δ→+0∣−2+Δx+2∣=0\lim\limits_{\Delta\to+0}f(-2+\Delta x)=\lim\limits_{\Delta\to+0}|-2+\Delta x+2|=0

lim⁡Δ→−0f(−2+Δx)=lim⁡Δ→−0∣−2+Δx+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)=4f(-2)=4. Therefore, the function is not continuous at x=−2x=-2.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS