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Mathematics 46 Online
OpenStudy (anonymous):

let f(x)\[\left[\begin{matrix}c^2\cos(x)-C & if x \le0 \\ c^2x^3 & if 0

OpenStudy (freckles):

we will have to consider left limit of 0=right limit of 0 and left limit of 1=right limit of 1

OpenStudy (freckles):

\[c^2\cos(0)-C=c^2(0)^3 \\ c^2(1)^3=3c(1)-2\]

OpenStudy (anonymous):

okay...

OpenStudy (freckles):

have you found C and c yet?

OpenStudy (anonymous):

no, not sure how to...

OpenStudy (freckles):

so you don't know how to simplify the equations I gave ?

OpenStudy (freckles):

cos(0)=1 0^3=0 1^3=1

OpenStudy (freckles):

\[c^2-C=0 \\ c^2=3c-2\]

OpenStudy (freckles):

you have a system of equations to solve

OpenStudy (freckles):

find lower case c from the second equation then you can go back in find big C from the first equation

OpenStudy (freckles):

don't be afraid of the second equation it is just a quadratic

OpenStudy (freckles):

you solved quadratics before

OpenStudy (freckles):

I think

OpenStudy (freckles):

I assumed

OpenStudy (freckles):

\[Ax^2+Bx+C=0 \\ x=\frac{-B \pm \sqrt{B^2-4AC}}{2A}\]

OpenStudy (anonymous):

does this work: (c-2)(c-1)

OpenStudy (freckles):

or factoring

OpenStudy (freckles):

(c-2)(c-1)=0 when c=? or c=?

OpenStudy (anonymous):

c=2,1

OpenStudy (freckles):

it just says to find little c

OpenStudy (freckles):

and there are two possible little c's

OpenStudy (freckles):

I wonder if they meant that big C to be a little C

OpenStudy (anonymous):

idk

OpenStudy (anonymous):

thanks for your help! appreciated it!

OpenStudy (freckles):

I want to add one more thing since your question asks for little c and there is a C and a c I kinda want to assume they meant C as c

OpenStudy (freckles):

and if so ... \[c^2-C=0 \\ \text{ is } c^2-c=0 \\ c(c-1)=0 \\ \text{ which means } c=0 or c=1 \] but the intersection of both of these conclusions gives that c has to be 1

OpenStudy (freckles):

the thing about the question is weird because you cannot find c such that it makes the function continuous you have to find (c,C)

OpenStudy (freckles):

to make the function continuous but it doesn't say that

OpenStudy (freckles):

so that is why I'm drawing the conclusion there is a type-0

OpenStudy (freckles):

either in them asking to find the possible values (c,C) or that they didn't mean to put a big C at all

OpenStudy (freckles):

you get what I'm saying @Brittni0605

OpenStudy (freckles):

wrong user

OpenStudy (freckles):

you get what I'm saying @esam2

OpenStudy (anonymous):

yes, i get what you mean.. i may need to check it with my teacher

OpenStudy (freckles):

we will do both cases: (c,C)=(c,c^2)=(2,4) or (1,1) --- or if they meant C=c then the answer is just c=1

OpenStudy (anonymous):

what happen to x=2?

OpenStudy (freckles):

x=2?

OpenStudy (freckles):

that wasn't a endpoint so we didn't need to consider it for making the function continuous since the piece that x=2 is a part of is continuous itself at x=2 from left and right

OpenStudy (anonymous):

oooh.... smart human!

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