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Its #4 only
For #4, draw a vertical line through -4 on the x axis where does this vertical line cross the red f(x) curve?
@ (-2,1) and (2,2)
oh wait, sorry those are the points of intersection of the two functions, one sec.
does this have to do with the vertical line test?
no, I'm simply asking for the value of f(-4)
Ok, could you help me with the domain and range part of the problem please.?
so you figured out that part already? the f(-4) part?
that wasnt the part I needed help with.. just included it because the graph is required in order to understand the rest of the problem..
the domain is the set of allowed inputs, in other words, the set of allowed x values
the red f(x) function stretches from x = -4 to x = 4 the function is not defined for other x values (like x = 10)
so that's why the domain is \[\Large -4 \le x \le 4\] in interval notation, the domain is written as `[-4, 4]` Notice the use of square brackets to mean include the endpoints
what would be the domain of g(x) ?
ok, I'll solve that now, could you explain to me when I include brackets and parenthesis when writing these things..?
The domain of g(x) is [-4,3]
the parenthesis are used to exclude the endpoint that's when you come across an endpoint with an open circle like so |dw:1440727842554:dw|
think of it as like a road that isn't finished. There is a pot hole at the open circle. It is NOT included in part of the road because you can't drive on it
So when there is a circle, I use brackets?
oh wait parenthesis nvm
ok thanks, could you help me with the range too plz?
the range is the set of possible outputs of a function put another way, the range is the set of possible y values that could come out
look at the red curve f(x) the lowest it goes is to y = -2 the highest it goes is to y = 3 the range of f(x) is \[\Large -2 \le y \le 3\] in interval notation the range of f(x) is `[-2,3]` I'll let you do g(x)
ok, is my domain for g(x) correct?
what did you say for the domain of g(x)?
oh nvm, you wrote `The domain of g(x) is [-4,3]`
yeah that is correct
ok, give me a sec for the other one...
1/2 is less than or equal to y is less than or equal to 4
I don't agree with the 1/2 part
oh wait, nvm
each tick is 1
its hard to tell
1/2 is a good estimate
what would the range be in interval notation?
ok, cool, Could you explain to me when I use less than and less than or equal to, is there a clear way of knowing?
it depends on if you have open or closed circles |dw:1440728925694:dw|
when you include a point, you say "or equal to" eg: \[\Large x \le 5\] we're including 5
saying `x > 5` we mean everything larger than 5 and we exclude 5 from the set
ok cool, thanks!
a) f(-4)=-2, g(3)=4. b) The values of x where f(x)=g(x) are x=-2 and x=2. c) An estimation of the solution of the equation f(x)=-1, is x=3. d) The interval on what f is decreasing is [0,4]. e) The domain of f(x) is [-4,4], the range of f(x) is [-2,3]. f) The domain of g(x) is [-4,3], the range of g(x) is [1/2,4].
@jim_thompson5910 does anything look incorrect?
`c) An estimation of the solution of the equation f(x)=-1, is x=3.` there's one other solution
`d) The interval on what f is decreasing is [0,4].` f is not decreasing when x = 0 or at the endpoint x = 4 so you'll need to exclude the endpoints and say \(\Large (0,4)\)
ok cool, does everything else look ok?
everything else is perfect
ok, and for c, how would I find that other point?
draw a horizontal line through y = -1 and see where it crosses the red f(x) function
you already found (3,-1)
actually I think you meant to say (-3,-1)
if f(x) = -1, then x = -3 or x = ???
so its x=-3 and 4?
yes if f(x) = -1, then x = -3 or x = 4
ok great, well thanks again for your wonderful help. Saved me alot of time, hope you have an awesome rest of your day!
glad to be of help