let f(x)\[\left[\begin{matrix}c^2\cos(x)-C & if x \le0 \\ c^2x^3 & if 0
we will have to consider left limit of 0=right limit of 0 and left limit of 1=right limit of 1
\[c^2\cos(0)-C=c^2(0)^3 \\ c^2(1)^3=3c(1)-2\]
okay...
have you found C and c yet?
no, not sure how to...
so you don't know how to simplify the equations I gave ?
cos(0)=1 0^3=0 1^3=1
\[c^2-C=0 \\ c^2=3c-2\]
you have a system of equations to solve
find lower case c from the second equation then you can go back in find big C from the first equation
don't be afraid of the second equation it is just a quadratic
you solved quadratics before
I think
I assumed
\[Ax^2+Bx+C=0 \\ x=\frac{-B \pm \sqrt{B^2-4AC}}{2A}\]
does this work: (c-2)(c-1)
or factoring
(c-2)(c-1)=0 when c=? or c=?
c=2,1
it just says to find little c
and there are two possible little c's
I wonder if they meant that big C to be a little C
idk
thanks for your help! appreciated it!
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
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
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)
to make the function continuous but it doesn't say that
so that is why I'm drawing the conclusion there is a type-0
either in them asking to find the possible values (c,C) or that they didn't mean to put a big C at all
you get what I'm saying @Brittni0605
wrong user
you get what I'm saying @esam2
yes, i get what you mean.. i may need to check it with my teacher
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
what happen to x=2?
x=2?
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
oooh.... smart human!
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