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?
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.
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.
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)
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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.
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I love your introduction of your videos
Nice example and work through. What type of chalkboard you use?
The first one that I see this kind of function . Thank you very much. Bye.
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 🙃
Keep on doing good work...u make our life
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?
Excellent lecture Sir. Thanks 🙏
Coolest teacher ever 🎉❤
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.
Vibrant video
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thank you so much for the help king! loved it
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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.
This man istg saves ass
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
Thanks for ever ❤
Oh, you made it continuous but not differentiable.
@awrRoman25 Yes, I said that.
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!