continuious means they both approach the same number as they approach 1 basically, find the vallues of q in terms of p for which both of them equal 1 so 1+2p(1-1)+(1-1)^2=1 (this simplifies to 1=1 so don't mess with the p) q(1)+p=1, q+p=1, q=1-p
the value of q that will make it continous at x=1 is 1-p
B. find the values of p and q such that their derivitives are equal at x=1
first one's derivitive: 2px+2x-2, at x=1, 2p+2-2=2p
2nd one's derivitive: q
so
2p=q from the value in part a q=1-p 2p=1-p add p to both sides 3p=1 divide by 3 p=1/3 q=1-1/3 q=2/3
the function is
1+(2/3)(x-1)+(x-1)^2 for x≤1 and (2/3)x+(1/3) for x>1
C. take the derivitive twice to get 0 and 0 both same deriritive
find if they have the same value well, 0=0
yes
A. q=1-p B. p=1/3, q=2/3 C. yes, both have value of 0 and change of slope is 0
