Question

Determine values of k, and m such that the following function g(x) is continuous and differentiable at all points.

(3x+k if x < 0)
g(x) = ( )
( e^mx if x_>0 )

Kind regards thanks...Please show work

Answer 1

Well, it looks like we could possibly have a jump discontinuity at x=0, unless we choose k in a way to correct this...

Answer 2

okay, how do you determine the jump discontinuouty

Answer 3

well, if x=0, then\[e^{m(0)}=1\]This tells us that we want k such that \[\lim_{x \rightarrow 0}3x+k=1\]This will preserve continuity at x=0.

Answer 4

yes

Answer 5

actually I made an error since we can only compute the left side limit for that function (x<0)

Answer 6

But, if the left limit and the right limit\[\lim_{x \rightarrow 0^+}e^{mx}=1\]are the same, and g(0)=1, then the function is continuous at x=0.

Answer 7

Once we fix this one trouble point, we see that the function will be continuous and differentiable everywhere.

Answer 8

Can you take it from here?

Answer 9

I am rather ignorant of how to solve this, i confess. Can you perhaps outline the steps and I will be able to do it then. ps. I have never solved this type of problem before although i have one in my notes, which are a bit vague.

Answer 10

you need to solve the limit\[\lim_{x \rightarrow 0^-}3x+k=1\]for k. Not sure how?

Answer 11

We can also write this as:\[\lim_{x \rightarrow 0^-}3x +\lim_{x \rightarrow 0^-}k=1\]

Answer 12

Not sure. Why do you equate it to 1, and why is limit at 0?

Answer 13

ok. our trouble will be at x=0 because we have a piecewise function that changes at x=0.

Answer 14

okay

Answer 15

We can plug x=0 into the second piece of the function (since x=0 is defined there but not defined for 3x+k). This will tell us that g(0)=1.

Answer 16

okay, so the limit of the first term is k=1

Answer 17

Yes. Because 3x approaches zero as x approaches 0 from the left. So, in order to make sure that the limit of 3x+k=1 as x approaches zero, we need to have k=1.

Answer 18

okay, do we take limit of second function too?

Answer 19

So what did we do here. We picked k=1 to ensure that the limit as x==>0 of g(x)=1 and we know that g(0)=1, therefore the function is continuous at x=0.

Answer 20

we are done, actually. Because no matter what m is equal to, this function will be continuous at x=0 and in fact, continuous and differentiable everywhere.

Answer 21

make sense?

Answer 22

okay, what about the x value in e^mx (is that 0 as well)

Answer 23

x is just a variable. It can take any value of x equal to or greater than zero.

Answer 24

your asked to pick constants m and k such that this function g(x) is continuous and differentiable everywhere. we are not "solving for x" here

Answer 25

you're*

Answer 26

so the value 0 is taken from the function g(0) and this value is used i both functions ?

Answer 27

I chose to examine the behaviour of g(x) around x=0 because the function is defined totally differently at x=0 (it is a piecewise function).

Answer 28

i have never come across that term piecewise, but I will revise it for sure

Answer 29

It just means you have two or more "functions" which together define a single function g(x). The range of inputs is different for each function but overlaps at x=0.

Answer 30

I suggest you review the definition of "continuous function". What criteria need to be met for a function to be continuous at a given x...

Answer 31

Right, so you look at the conditions -> x<0 and x >_ 0 and see where it overlaps. The overlap is the value for x shich is then inputted to the function.

Answer 32

In a nutshell, yeah. Just make sure you input it in the right piece of the function. You can only input x=0 into the e^(mx) part. x=0 is not an allowable input for the 3x+k part.

Answer 33

One other question. You solved e^mx first and this gave you a value of 1. As a general tule fdo you always solve the second equation 1st.

Answer 34

No. If the restrictions on 3x+k where x less than or EQUAL to zero and e^(mx) were x greater than zero then we would have to start at the first equation.

Answer 35

So the EQUAL to restriction is what determines the order?

Answer 36

yeah. We need to know what g(0) is. We can only find that out using the equation that includes zero as an input. There is just no other way to do it.

Answer 37

okay,excellent. You have been amazing tonight. Your explanation is brilliant and your patience outstanding!

Answer 38

thank you :)

Answer 39

Thank you very much. My heartfelt gratitude. take care buddy!

Answer 40

you too!