$$\begin{cases} kx + 8 & x < -6\\ -9x + k & x \geq -6 \end{cases}$$
When I did my work
$$kx+8 = -9x+k\\ k(-6)+8 = -9(-6)+k\\ k(-6)+8 = 54+k\\ k(-6) = 46+k$$
How do I go from here?
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kx + 8 and -9x + k For what values of $x$? – 5xum Aug 28 '15 at 07:06
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kx + 8 when x < -6 and -9x + k when x > or equal to -6. – beginningprogrammer Aug 28 '15 at 07:08
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Please edit your question with this information. It's bad form if half the question is in the comments. – 5xum Aug 28 '15 at 07:13
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My apologies, I'm new to this site. I added that to my problem. – beginningprogrammer Aug 28 '15 at 07:27
1 Answers
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You have an equation for $k$. The equation is
$$-6k = 46 + k$$
You can solve this equation for $k$ and you will get your solution.
The more important thing is to know where your equation came from. The thing is that your function is equal to $kx+8$ for $x<6$, meaning that the limit
$$\lim_{x\uparrow -6} f(x)$$
is equal to $$\lim_{x\uparrow -6} kx+8 = k\cdot (-6) + 8$$
while the upper limit $$\lim_{x\downarrow -6} f(x)$$
is (because $f(x)=-9x+k$ for $k>-6$) equal to
$$\lim_{x\downarrow -6} -9x+k = -9\cdot(-6)+k$$
and the function can only be continuous if these limits are equal.
5xum
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1Sorry I have brain lapses, you put the -6 infront which made my brain start clicking again. -9.2 – beginningprogrammer Aug 28 '15 at 07:37
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Ok Thank you. I had the problem correct when I was randomly inputting items until I was close, but never put it into fraction form, that is why the website marked it as wrong. Also instead of dividing 7 by 46 I divided 5 by 46. Simple addition error on my part. – beginningprogrammer Aug 28 '15 at 07:44
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@beginningprogrammer Yeah, from $-6k=46+k$, you get $-7k=46$, since the plus flips to a minus. – 5xum Aug 28 '15 at 07:46
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