Question

Find the value of c and d so that f(x) is both continuous and differentiable.
f(x)= e^x +x^2 +c x0

Answer 1

I know that for the functions will need to be equal at any possible discontinuity points...
and the limits of each function and of the derivatives of the functions must be equal... I'm just not sure how to start and I would love some help!

Answer 2

"the limits of each function must be equal "


so start by finding the limit with x->0- and x->0+ of f(x)

Answer 3

\(\large \lim \limits_{x \to 0^- }f(x) =\lim \limits_{x \to 0^+ }f(x) =f(0)\)

this is true because f(x) is continuous

can you find those 2 limits ?

Answer 4

Ok I'm a little confused as to how to find the limits when the equations have c and d in them... x-> 0- would use the equation e^x +x^2 +c
and x->0+ would use dx +2
so I need to find c and d that would make them equal?

Answer 5

so for them to be continuous, e^x + x^2 +c =0 and dx +2=0 and I just have to find out what they are then?
ok I solved this and I got
d= -2/x and c= -e^x - x^2
can anyone tell me if this sounds right?
I used the substitution method and I think it is correct

Answer 6

oh, but that's just for continuity... what do I do for differentiability?

Answer 7

Hmm maybe I'm missing something here, this problem looks a lot simpler than what you have.

Answer 8

Hi, thank you so much if you can offer some help... take your time, I know you will want to see what I have written so far
so I think I have the values of c and d for the equations to be continuous, but I'm not sure what to do to make them differentiable
Also, the values that I got for c and d have variables, should they be like that?

Answer 9

\[\Large\sf \lim_{x\to 0^-}e^x+x^2+c=\lim_{x\to0^+}dx+2\]"At" x=0 we're getting,\[\Large\sf e^0+c=d\cdot 0+2\]Which simply gives us c=1, yes? :o

Answer 10

does it? my problem is that I tend to overthink things and then I hit a standstill because I always over-complicate the problem.
That's probably the case here, too... so how would you suggest solving it?

Answer 11

*headdesk* yes, yes it does give us c=1. I was trying to solve this by looking at a sort-of-similar problem on yahoo answers, obviously my solution was way to complicated...

Answer 12

So umm, I think this just works out similar to a problem we had done like last week or so.
For differentiability we need a nice smooth connection: The tangent lines must be approaching the same value at the joint.

So the lim of the derivative function from the left and right must match.
And then we plug in x=0 again to get d.

Answer 13

Somethingggg like this, yes?
\[\Large\sf \lim_{x\to0^-}e^x+2x=\lim_{x\to0^+}d\]

Answer 14

ok right, I know you helped me with the same type of problem, but I was thrown off by the addition of unknowns like c and d
ok so what you just wrote would be the limits of the derivatives of the equations... makes sense

Answer 15

Then just plug in x=0 to find your d, yes?

Hmm this problem worked out really nice and simple.
See how your d value disappeared in the first equation,
and the c value disappeared in the second.

Sometimes you'll come across problems where they don't disappear like that, and you're left with a system of equations so you have to do a little bit more work :p

Answer 16

I would plug in x=0 into... lim x->0- e^x 2x = limx->0+ d?
so would I have e^0 +2(0)= d?
d=1 ?

Answer 17

Mmm yah that looks right! :o

Answer 18

so c=1 and d=1 for the equations to be continuous and differentiable!
thank you so much for your help!