find value k making h(x) continuous on [0,5]:
h(x)= e^kx (0 is less than or equal to x is less than 2)
x+11 (2 is less than or equal to x is less than or equal to 5)

Question

find value k making h(x) continuous on [0,5]:
h(x)= e^kx (0 is less than or equal to x is less than 2)
          x+11 (2 is less than or equal to x is less than or equal to 5)

zepdrix · Answer

$$\Large\rm h(x)=\cases{e^{kx}, &$0\le x\le2$\ \rm x+11, &$2\le x\le5$}$$So this is our function huh? :d

anonymous · Answer

yes! (sorry for the late reply)

zepdrix · Answer

For continuity, we'll want the pieces to connect at x=2.
We look at the limit from the left and right sides of the function.

zepdrix · Answer

$$\Large\rm \lim_{x\to2^-}\color{orangered}{h(x)}=\lim_{x\to2^+}\color{orangered}{h(x)}$$We need these limits to agree.

zepdrix · Answer

$$\Large\rm \lim_{x\to2^-}\color{orangered}{e^{kx}}=\lim_{x\to2^+}\color{orangered}{x+11}$$Do you understand how I was able to plug these in for h(x)?

anonymous · Answer

yes!

zepdrix · Answer

Since we have no discontinuity (problem) at x=2, we can apply these limits by simply plugging in x=2 for each side.$$\Large\rm e^{2k}=2+11$$

zepdrix · Answer

And then solve for k.
Do you remember how to use your log rules? :O
We'll need one of them here.

anonymous · Answer

ohh okay thank you! and i don't remember my log rules.. :( but i can search them up!

zepdrix · Answer

This is the one that you'll want to remember:$$\Large\rm \ln(a^{\color{orangered}{b}})=\color{orangered}{b}\ln(a)$$

zepdrix · Answer

Taking the log of each side will allow you to get the k OUT of the exponent position.

anonymous · Answer

ohh that's right! wow thank you.

anonymous · Answer

so when i log the 2 + 11 side, do i just calculate those values on my calculator?

zepdrix · Answer

Mmmmm yah, or just leave it as a log. That probably is better actually.

Hmm I'm trying to make sure I didn't make a boo boo somewhere, it's not graphing correctly for me.

zepdrix · Answer

Ahhh nevermind it seems to be working out ok >.<
I was just being silly I think.

zepdrix · Answer

What'd you come up with?

anonymous · Answer

i haven't done the work yet! i'm working on another problem at the moment; once i'm finished i will post the answer here, though, if you'd like to check! thank you again

anonymous · Answer

i haven't simplified my work yet, but i got (ln2 + ln11)/2e ... i feel like this is wrong /:

anonymous · Answer

figured it out! i think! haha i got ln(13)/2

zepdrix · Answer

Ah there we go \c:/ yay team!

anonymous · Answer

thank you!!