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F:f(x)=

  • $3x +2$ at $-3<=x<2$

  • $X^2+4$ at $2<=x<=5$

Okay may any one please explain this from a to z.. I have been trying to solve it since a long time and I can't explain in details.

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    Just insert $x=2$ into the two functions and verify whether the values coincide: This is the case here, hence $f(x)$ is continous in $[-3,5]$ – 2017-01-16
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    @Peter what? How? – 2017-01-16
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    You have to verify whether the limit for $x\rightarrow 2$ exists. To do that, you have to compare the limits from the left and the right – 2017-01-16
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    @Peter look I have an exam tomorrow, and this is a main example in my book it is already solved but i don't know how did they solve it. – 2017-01-16
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    I gave a detailed answer. If you still need explanations, please tell me. – 2017-01-16

1 Answers 1

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First of all, polynomials are continous at every position $x_0$. So we only have to consider the position $x_0=2$.

We must check whether $$\lim_{x\rightarrow 2} f(x)$$

exists. For $x<2$ , we have $f(x)=3x+2$. Hence, we have

$$\lim_{x\rightarrow 2-0} f(x)=\lim_{x\rightarrow 2} 3x+2=8$$

"$2-0$" means that we approach from the left.

For $x\ge 2$, we have $f(x)=x^2+4$ , hence $$\lim_{x\rightarrow 2+0} f(x)=\lim_{x\rightarrow 2} x^2+4=8$$

"$2+0$" means that we approach from the right.

Finally, we have $f(2)=8$, so the limit exists AND it coincides with the value of $f(x)$ at $x_0=2$.

We can conclude that $f(x)$ is continous in $(-3,5)$

For $x=-3$, $f(x)$ is only right-continous because there is no limit from the left and for $x=5$, $f(x)$ is only left-continous because the is no limit from the right.

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    But by this we proved that $f$ is continuous at $x=2$ how did we concluded that interval? Shouldnt i prove that it is continuous on the right of$x=3$ then the left of $x=5$ to get that interval? And why we chose that interval? Not [2,5] or] -3,2[ – 2017-01-16
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    Yes, you are right in some sense. At $x=-3$, $f(x)$ is only right-continous and for $x=5$, $f(x)$ is only left-continous. I will fix my answer ... – 2017-01-16
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    I understood this part but my book says, we should discuss 3 things -1 Continuity of the two intervals ]-3,2[ and ]2,5[ (why they are opened?) – 2017-01-16
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    You made the second step only, and I already mentioned the 3rd step. – 2017-01-16
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    Without calculating the limits, we only know that $f(x)$ is continous for $x<2$ and $x>2$. Therefore the open intervals. To check continuity for $x_0=2$ , we need the limits because it is not clear in advance that they coincide. – 2017-01-16
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    What is the $3rd$ step I missed ? – 2017-01-16
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    And we closed the interval to say that f(x) is continuous at (-3,5) why? – 2017-01-16
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    3rd step is Continuity from right at x=-3 and the left at x=5 – 2017-01-16
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    You can use that polynomials are continous, hence left-/rightcontinous, if you deal with such functions. If you do want it, you can argue that $$\lim_{x\rightarrow 5-0} f(x)=\lim_{x\rightarrow 5} x^2+4=29=f(5)$$ and analogue for $x=-3$ – 2017-01-16
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    okay thank you so much I understood it now! – 2017-01-16
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    Good luck for your exam! – 2017-01-16
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    thanks but I still have some other questions be there for me you are awesome teacher! – 2017-01-16
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    Thanks, but for more questions you should ask another question. Otherwise the comment-list gets too long ... – 2017-01-16
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    i know :D thanks – 2017-01-16