For the function defined by... Precalculus question?
For the function defined by: f(x) = \[f(x) = \left[\begin{matrix}x^2 & x \le 1 \\ 2x+1 & x<1\end{matrix}\right] \]
Pretend the brackets look like these { }
Part 1: Evaluate f(0). Part 2: Graph f(x).
can you check your function again and make sure the inequalities are posted correctly?
there is some ambiguity in the question because f(0)=0 by the first equation. also f(0)=2*0+1=1 by the second eq.
Oops - the last inequality (bottom right) should be x>1
ok cool
Also note that the brackets on the end are actually squiggly lines { }
so, if x = 0, which function do we use?
Do you know how to do it?
yes
f(0) means evaluate the function when x = 0 now, this function has two rows. one row is when x >= 1 and the other row is when x < 1
f=0 would make all equations true except the bottom left one.. is that what you're asking?
yes, since x = 0 we only use the top row
so, if f(x) = x^2, when x = 0, what is f(x)?
I'm just not sure how to do it since there are four different functions... is it a matrix?
there are only two functions
good, so f(0) = 0
this is how the function works
It would equal 0
this means that f(x) = x^2 when x <=1
the second row means that f(x) = 2x + 1 when x > 1 make sense?
Hold on, let me look at this
Not quite sure what you mean
Why are we only using the top two equations?
there are two equations
the x <=1 part is not an equation, it tells us to use x^2 only when x <=1
since 0 is less than 1, we use x^2
make sense?
Oh I see
so, for the graphing part:
That clears things up!
for graphing, all you do is graph x^2 but only for x <=1
so your graph will look like this
So we would use the first one because x is less than or equal to 1?
yes
Ah okay
Thanks so much for your help! I think I got it from here
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