Question

How do I solve this problem

Answer 1

For part a) Set 2 functions equal when x=1
$q+p = 1 \\ q = 1-p$

For part b) Set 2 function derivatives equal when x=1
$q=2p$
Sub in equation from part a) to solve for p,q
$1-p=2p \\ p = \frac{1}{3} , q = \frac{2}{3}$

part c) Take 2nd derivatives. When x<=1, f''(x) = 2, when x>1, f''(x) = 0
Therefore f''(x) is not continuous.

Answer 2

A.

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