Question

Determine the values of the constants B and C so that the function given below is differentiable.
f(x){ 11x x<=1
{ Bx^2+Cx 1 14 years ago

Answer 1

gotta match em up

Answer 2

the derivative from the left is 11

Answer 3

derivative from the right is
\[f'(x)=2bx+c\]

Answer 4

so you need two things. you need the functions to agree at 1 that is you need
\[11\times 1 = B\times 1^2+C\times 1\] and you need
\[2B \times 1 +c=11\]

Answer 5

got two equations with two unknowns. first one is
\[B+C=11\] second one is \[2B+C=11\]

Answer 6

ok

Answer 7

hold the phone maybe i made a mistake. i get B = 0 and C = 11, but that just gives what you started with. just the same i think i am right

Answer 8

derivatives have to match, and the function has to be continuous there;.

Answer 9

so for sure you know
\[11\times 1 = B\times 1^2+C\times 1\]
i.e.
\[B+C=1\] no way of avoiding that

Answer 10

if it is not continuous then it is not differentiable that is for sure

Answer 11

and also the derivatives have to be the same from the left and from the right

Answer 12

derivative from the right is 11 for sure

Answer 13

going over it now

Answer 14

and derivative from the left is
\[2Bx+C\] also for sure

Answer 15

yes

Answer 16

ok so i get
\[2B+C=11\] and \[B+C=11\] forcing B = 0 and C = 11 and so it must be
\[f(x)=11x\] all the way

Answer 17

i am sticking to that

Answer 18

that is one of the answers!

Answer 19

makes sense to me too!

Answer 20

3 more and Im done!