Question

https://www.sophia.org/download/ckeditor%2Fpictures/114112/data/content/sophiatutorials_tdifcont05.jpg?

Answer 1

let me make new notations for convience, if you don't mind

Answer 2

\(\large\color{black}{ \displaystyle f_{\text{#}1}=ax-1,{~~~}(-\infty,b) }\)
\(\large\color{black}{ \displaystyle f_{\text{#}2}=x^2,{~~~~~~~~~~}[b,+\infty) }\)

Answer 3

For the function to be continuous,

\(\large\color{black}{ \displaystyle f_{\text{#}1}(b)=f_{\text{#}2}(b) }\)



For the function to be differentiable,

\(\large\color{black}{ \displaystyle f_{\text{#}1}'(b)=f_{\text{#}2}'(b) }\)

Answer 4

So for the function to be continuous:
\(\large\color{black}{ \displaystyle a\times \color{red}{b}-1=\color{red}{b}^2 }\)

For the function to be differentiable,

\(\large\color{black}{ \displaystyle a=2\color{red}{b} }\)

Answer 5

there is the system of equations for you to solve.
Hope this helps, and if you have any questions, ask

Answer 6

sorry i'm still kind of confused

Answer 7

Ok, what exactly confuses you?

Answer 8

f′#1(b)=f′#2(b)

Answer 9

that is read as:

the first part of the function (the ax-1), evaluated at x=b,

is equal to

the second part of the function (the x²), evaluated at x=b.

Answer 10

and the reason I set this up, is because for the function to be
continuous its two peaces must be connected and the only way
to get this is to set: `f#1(b)=f#2(b)`

Answer 11

Oh, and with primes, I am setting their slopes equal to each other

Answer 12

the slope of the first peace and the second peace at x=b must be equal (or else the function is not differentiable).

Answer 13

f′#1(b)=f′#2(b)

is read as:

the derivative of the first peace evaluated at x=b,
is equal to
the derivative of the second peace evaluated at x=b.

Answer 14

is the answer going to be (D), i don't know if i did it right

Answer 15

Do you understand how I get the system of equations:

\(\large\color{black}{ \displaystyle a b-1=b^2 }\)

\(\large\color{black}{ \displaystyle a=2b}\)

then I solve:

\(\color{black}{ \displaystyle (2b)b-1=b^2 }\) >> \(\color{black}{ \displaystyle 2b^2-1=b^2 }\) >> \(\color{black}{ \displaystyle b^2=1 }\) >> \(\color{black}{ \displaystyle b=\pm1 }\)

\(\color{black}{ \displaystyle a=\pm2}\)

Answer 16

ohh okay

Answer 17

so technically there is more than one answer, really.
But, I guess that a square root is presumed to be positive in this case, and thus b=+1, and consequentially a=+2.

Answer 18

so then (A) a=2, b=1 or a=-2, b=-1

Answer 19

Oh, you have two options, then you don't have to presume a positive root.

Answer 20

so b=±1,
when b=-1, a=-2,
when b=1, a=2.

Answer 21

yes, option A is right

Answer 22

kay, thank you for your help

Answer 23

yw