For f(x) to be continuous everywhere, the lim x-->-2 must exist aka lim x-->-2 (ax^2+bx)= lim x-->-2 (ax^3-10b)= (-7x-2)=14. Therefore, a(-2)^2+b(-2)=14=a(-2)^3-10b by plugging in x=-2 or 4a-2b=14 and -8a-10b=14 as well. Dividing the second equation by -2 gives 4a-2b=14 and 4a+5b=-7 and subtracting the second from the first gives (4a-2b)-(4a+5b)=14-(-7)--> -7b=21 or b=-3. Then plugging b=-3 into equation 1 gives 4a-2(-3)=14 or 4a-(-6)=14 or 4a=8 so a=2. Therefore (a,b)=(2,-3) for f(x) to be continuous everywhere.
I really like this type of exercise! For fun, I also tried making the first derivative continuous (removing the middle definition, that is redundant), but the solution was the identically zero function 🙃
When studying computer graphics many years ago, I recall the importance of C2 continuity for creating smooth curves and surfaces. Perhaps you could extend this problem to a C2 function?
If the question was phrased “Find the values of a and b for which f is differentiable everywhere” we were gonna different values of a and b which satisfy continuity and differentiability.
Why the limes evaluation? Just insert the value at "x = -2" into the two formulars together with the x and solve the simple linear two equations! The derivatives we would need if we would want the crossover to become smooth in the derivatives. But that would require a few more degrees of freedom for our functions.
for f'(x) to be continuous, all we need are different values of a and b. f'(x)={2ax+b, x-2 -4a+b=-7 12a=-7 a=-7/12 b=-14/3 f'(x)={-7x/6-14/3, x-2 f(x)={(-7/12)x^2-(14/3)x, x-2
![lim [5th-root(x) - 2]/[(x-32)] as x goes to 32](http://i.ytimg.com/vi/_6XobtpX5cM/mqdefault.jpg)
![lim [5th-root(x) - 2]/[(x-32)] as x goes to 32](/img/tr.png)
I LOVE YOU PRIME NEWTONS!! i got a floor functions question on my maths test and knew exactly how to solve it
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Thank you, Newton, I was really struggling with these types of problems. Big respect ✊🏽
For f(x) to be continuous everywhere, the lim x-->-2 must exist aka lim x-->-2 (ax^2+bx)= lim x-->-2 (ax^3-10b)= (-7x-2)=14. Therefore, a(-2)^2+b(-2)=14=a(-2)^3-10b by plugging in x=-2 or 4a-2b=14 and -8a-10b=14 as well. Dividing the second equation by -2 gives 4a-2b=14 and 4a+5b=-7 and subtracting the second from the first gives (4a-2b)-(4a+5b)=14-(-7)--> -7b=21 or b=-3. Then plugging b=-3 into equation 1 gives 4a-2(-3)=14 or 4a-(-6)=14 or 4a=8 so a=2. Therefore (a,b)=(2,-3) for f(x) to be continuous everywhere.
If I were God I would only add you more years to live forever and ever more, you have really made me understand limits in less than 6 minutes.
More Calc 2 content please🥺
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I appreciate you more live long life I'm from ethiopia 🇪🇹 in oromoo regions
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@@GiftMlinde Kenya?
I really like this type of exercise! For fun, I also tried making the first derivative continuous (removing the middle definition, that is redundant), but the solution was the identically zero function 🙃
Excellent lecture Sir. Thanks 🙏
When studying computer graphics many years ago, I recall the importance of C2 continuity for creating smooth curves and surfaces. Perhaps you could extend this problem to a C2 function?
Keep on doing good work...u make our life
Thanks for making understand how to solve this kind of problems newton. You are the best ❤❤
I love your introduction of your videos
If the question was phrased “Find the values of a and b for which f is differentiable everywhere” we were gonna different values of a and b which satisfy continuity and differentiability.
Prime Newtons has it all! 😊
He's the OG
Thank you, thank you, thank you (south africa)
Nice example and work through. What type of chalkboard you use?
Why the limes evaluation? Just insert the value at "x = -2" into the two formulars together with the x and solve the simple linear two equations! The derivatives we would need if we would want the crossover to become smooth in the derivatives. But that would require a few more degrees of freedom for our functions.
Oh, you made it continuous but not differentiable.
@awrRoman25 Yes, I said that.
This was really helpful ❤️ thanks alot❤
Coolest teacher ever 🎉❤
The first one that I see this kind of function . Thank you very much. Bye.
a(-2)^2+b(-2)=-7(-2)=a(-2)^3-10b
4a-2b=14; -8a-10b=14
12a+8b=0
a=(-2/3)b
(-14/3)b=14
b=-3
a=2
f(x)={2x^2-3x, x-2
f'(x)={4x-3, x-2
although the function is now continuous at x=-2, its derivative isn't.
for f'(x) to be continuous, all we need are different values of a and b.
f'(x)={2ax+b, x-2
-4a+b=-7
12a=-7
a=-7/12
b=-14/3
f'(x)={-7x/6-14/3, x-2
f(x)={(-7/12)x^2-(14/3)x, x-2
Well explained
thank you so much for the help king! loved it
You are the best👌
Vibrant video
This man istg saves ass
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Thanks for ever ❤
b = -3 and a = 2
asnwer=2
OR, TO SAY IT DIFFERENTLY, WHEN YOU BUILD THE FUNCTION GRAPH, YOUR HAND SHOULD NOT BE INTERRUPTED DURING DRAWING IT!