find values c and d that make h(x) cont. on all real numbers?

Question

h(x)={ 2x if x< 1
cx^2 +d if 1 (less than or equal to) x (less than or equal to) 2
4x if x > 2

Answer 1

You need cx^2+d to become 2x=2 when x=1, and 4x=8 when x=2:
c+d=2
4c+d=8

Solving this, get: c=2, d=0

Answer 2

cx^2 + d = 2 when x = 1 hence c + d = 2
cx^2 + d = 8 when x = 2 hence 4c + d = 8
therefore c = 2 and d = 0

Answer 3

The possible continuities are in x = 1 and x = 2.

For x = 1, we want
[A] ... 2x = c x^2 + d
so
[A'] ... 2 = c + d

For x = 2, we want
[B] ... c x^2 + d = 4x
so
[B'] ... 4 c + d = 8

Now solve the system of equations [A'], [B']. Simplest is to subtract [A'] from [B'], you find

3 c = 6, so c = 2
and d = 0

Answer 4

for h to be continuous we have to see at 1 from left and right to be same and at 2 to be same

at 1 from the left the limit = 2x = 2
from the right the limit = c + d
so c+ d =2 ...2
at 2 limit from left = 4c + d
the value from rhs = 8
so 4c+ d = 8...2
we need to solve for c d from 1 and 2
subtract 1 from 2 3c = 6 so c = 2
so d= 2-2 = 0

Answer 5

c = 2
d = 0