Question

Determine values of k, l,m and n such that the following function g(x) is
continuous and differentiable at all points.

(2x^2 − n if x<−2 )
g(x)=(mx + l if −2<=x<2 )
(kx^2 + 1 if x>=2 )

Answer 1

any fixed ponts?
like 2 or -2?

Answer 2

no i put it up exactly as it is on the problem sheet im doing

Answer 3

4k= m , m= -8 so k=-2 . l=-9

Answer 4

n=1

Answer 5

how@manorajan.

Answer 6

how did you get those answers?

Answer 7

differentiate & set the derivatives equal of the secon two equations. this gives 4k=m at x=2. do this to the firat two at x=-2. now set the values of the functions at 2 & -2 same. this gives k,l.m.n

Answer 8

i.e 4k=m & from m=4x at x=-2 m=-8. so k must be -2. at x=-2 value of the last fn is -7. to be continuous the value of the second function must be the same. so -8x + l = -7

Answer 9

differentiate 2x^2-n & set the derivative equal of the secon two equations.
or differentiate 2x^2-n,and mx+l,and kx^2+1 & set the derivative equal of the secon two equations.?

Answer 10

4k+1=2m+l . you know k & m from above. this gives -7=-16+l. l must be -9, no?

Answer 11

you need to do it in pairs. the second two first, the first two next. then set values equal .. in pairs. at points 2 & -2. this gives you four equations & you have four unknowns. you can do it in any way of pairing as you like..

Answer 12

this is still new to me.. can you give me the first two derived equations and what you put them equal to plz?

Answer 13

ok i got it now,,,,the only thing im not sure of is when to use x=-2 and x=2?

Answer 14

right. i'll write it down fully, but remember that this is just a suggestion. Most places you need to jump out & find an easier way..

differentiate the last two equations and set them equal at x=2 :: 2.k.x=m; i.e 4k=m
Now, do this to the first pair, set them equal at x=-2:: 4x=m; i.e m=-8. this gives you m.
use this in 4k=m .. so k=-2.

now since the curve is continuous, the values of the functions at the endpoints are the same as the next fn after them.

for the first two at x-2:: 8-n=-2.m+l. m=-8, so 8-n=16+l. we dont know l yet.
from the second two at x=2::4.k+1=2.m+l. we know k&m. this gives us l=-9.

use this in 8-n=16+l. this gives 8-n=7. so n=1

Answer 15

ok i have it now thanks :)

Answer 16

like i said, if your cuve is continuous then the pieces must lie end to end. so from one function to the next.. at the end points (here these were 2 & -2) their values & their derivatives must be the same. this will give you sufficient equations for all your unknowns. all you need to do is solve them.

Answer 17

thanks