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If f(x) is continuous and differentiable and f(x) = ... then b? f(x) = {ax^4 + 5x, x< or = 2 {bx^2 - 3x, x> 2, then b = ? Help pls
1st set these two expressions equal to each other: \[\lim_{x \rightarrow 2^-}f(x)=\lim_{x \rightarrow 2^+}f(x)\]
2nd set these two expressions equal to each other: \[\lim_{x \rightarrow 2^-}f'(x)=\lim_{x \rightarrow 2^+}f'(x)\]
What should I do after that? Add those two equations together?
You will have a system of equations to solve just like from algebra class.
Well I got, 4a - 10 = 4b - 6 and then -32 + 5 = 4b - 3. Is that correct?
The first equation on the left hand side you have 4a-10 I think it should be 16a+10. What happen to a in the second expression?
\[a(2)^4+5(2)=b(2)^2-3(2) \text{ first equation }\]
\[4a(2)^3+5=2b(2)-3 \text{ second equation }\]
Wait isn't it a(-2)^4 + 5 (-2) since the limit is coming from the left side?
It isn't approaching -2. It is approaching 2.
You have the left is just approaching 2 from the left. You have the right is just approaching 2 from the right. Either way they are both still approaching 2 and not some other number.
Ahh ok I got it :) After I get these 2 equations then what should I do afterward?
Have you ever solved a system of two linear equations?
Yes, like 4 years ago haha.
16a + 10 4b - 6 32a + 5 4b - 3 ( Can't I just substitute it and then solve it) ?
You can put them both in ax+by=c form if you want. Like if we had something like 2x+3y=1 -2x+3y=1 This system is already set up for elimination. Add the equations together and solve for y first would be my strategy. 6y=2 y=2/6=1/3 Then plug into one of the other equations and solve for x.
Or this system: 3x-6y=2 x+y=2 This one isn't set up for elimination. But you could multiply the bottom equation by -3 so we would have it setup for elmination. 3x-6y=2 -3x-3y=-6 ----------------add -9y=-4 y=4/9 and then so on to find x.
\[a(2)^4+5(2)=b(2)^2-3(2) \text{ first equation }\] \[16a+10=4b-6\] \[16a-4b=-16 \text{ rewrote the equation 1} \]
Do you think you can rewrite equation 2 in that form?
32a - 4b = -8 ( 2nd equation right) right?
So we have 16a-4b=-16 32a-4b=-8
If we multiply the second or even the first (just not both of them) by -1, then we can set this up for elimination method.
I'm trying to eliminate b. (you could go for a but I prefer working with smaller numbers)
So -16a+4b=16 32a-4b=-8 ----------------add the equations together and what do we get?
I got the final answer of 6 for b. Is that right?
yayyy, you should just be my tutor haha. Thank you so much :D