Function Continuity(need help by Oct. 15)

[Gwennypoo]

abs. of (x-1) +2 for x<1.
f(x) { ax^2 +bx for x> or = to 1, where a and b are constants.


If a =2 and b=3, is f continuous for all x?? Justify.

Decribe all values of A and B for which f is continuous function.

For what values of A and B is f both continuous and differentiable?

i solved for a, stating that it's not continuous, but i'm not entirely sure that's right.. help would be great!!!

 

[arthur ohlsten]

a) abs [x-1] +2 for x<1
abs x-1 can be replaced by:
x-1 for 1<x
-x+1 for x<1

abs x-1 is a V shaped curve apex at 1,0

abs [x-1] +2 is a V shaped curve apex at x=1 y=2

abs[x-1]+2 is continuous
" a function is continuous if you can trace it with your finger without removing your finger from the paper."

abs[x-1] is not differentiable at x=1 because you get a different value for the derivative depending apon which way you approach the point.
you get dy/dx =1 to the right of x=1 and dy/dx =1 to the left of the point x=1

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rest to follow
Arthur

 

[arthur ohlsten]

Typo error on slope at x=1 when approached from the left dy/dx=-1

b)
y=ax^2 +bx x>1
for a=2
b=3

y=2x^2+3x
y=2 [x^2+3/2x] complete the square
y=2[x^2+3/2 x+ [3/4]^2] -2[3/4]^2
y=2[x+3/4]^2 -9/8 a parabola open up
vertex at -3/4 , -9/8

function is continuous
===========================================
y=ax^2 + bx
y=a[x^2+ b/a x] complete the square
y=a[x^2 + b/a x + [b/2a]^2] - a[b/2a]^2
y=a[x+b/2a]^2 - b^2/3a^2 a parabola open up for a>1 open down for a<1
vertex at -b/2a , [b^2/3a^2]

I believe this is always continuous , it is always a parabola
it is always differentiable as it is a smooth curve continuous

Arthur