Please forgive me if this question has been answered before, but I haven't found anything that seems to cover my problem (I am not a mathematician).

I have several mathematical problems that are like these ones (can't post screenshots unfortunately):

Problem 8 to 12 are all similar, problem 7 is more general, but yet I don't understand how it's done and don't find anything on the internet and I am running out of time (exam in T minus 26,5 hours).

The alleged solutions for problems 7-12 are:

  1. 7
  2. 197.35
  3. 126.90
  4. 394.13
  5. 57.39
  6. 135.84

Now my problem is, I don't know how they come up with these numbers and can't figure it out from the script.

Please help me out here! Thanks a lot!


Basically: The functions are continuous everywhere up to the point where the two domains meet, and you have to adjust the function so that the functions are also continuous at the "adjunction point".

For example given continuous functions $g(x),h(x)$ and $$f(x)=\begin{cases}g(x)& x>c\\h(x) & x\leq c \end{cases}$$

you need to check that the function is also continuous at $x=c$.

This, in turn, is nothing but checking that $$g(c)=h(c).$$

More formally the actual condition should be $$\lim_{x\searrow c}g(x)=\lim_{x\nearrow c}h(x)$$

  • $\begingroup$ Sorry, I just noticed that my answer is pretty similar to yours at some points, wasn't mean to copy. $\endgroup$ – Giulio Feb 5 '17 at 10:51
  • $\begingroup$ No problem at all :) $\endgroup$ – b00n heT Feb 5 '17 at 10:58

I give you a guideline for problem $7$, I'll leave to you to try the others.

First thing you have to notice is that if $x \lt1$, then $f(x) =g(x):= -x+k$ is continuous because sum of $2$ continuous functions. For $x\gt1$, $f(x)=h(x):=2x^2+4x$ which is continuous for the same reason.

So the only thing you have to check is whether $f(x)$ is continuous in $x=1$, which is the point where $g(x)$ and $h(x)$ meet.

To check it you can calculate the value of $k$ for which $g(1)=h(1)$, obtaining $k=7$. But this is not really formal, because $g(x)$ is not defined in $x=1$.

So you should do: $\lim_{x \to 1^-}g(x) =h(1)=6$, where $1^-$ means that you tend to $1$ from values of $x<1$, which gives you the desired result.


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