C3: Finding the range of a functionWatch

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Thread starter 6 years ago
#1
so there is this function f(x)=1+2arc tanx and im asked to find the range.

I don't understand how to do this as i wasn't given a domain, i found f^-1 (x) which was tan ((x-1)/2) but that didn't really help...any ideas?
0
6 years ago
#2
(Original post by Miss_Jazz)
so there is this function f(x)=1+2arc tanx and im asked to find the range.

I don't understand how to do this as i wasn't given a domain, i found f^-1 (x) which was tan ((x-1)/2) but that didn't really help...any ideas?
The range is just the maximum and minimum values that f(x) can equal. So a sketch would really help with finding the range of a function. For example, the range of sin(x) is 1>=f(x)>=-1, which equals the domain of arcsin(x).

You could try skecthing f(x), and think about the maximum and minimum output.
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Thread starter 6 years ago
#3
(Original post by hello calum)
The range is just the maximum and minimum values that f(x) can equal. So a sketch would really help with finding the range of a function. For example, the range of sin(x) is 1>=f(x)>=-1, which equals the domain of arcsin(x).

You could try skecthing f(x), and think about the maximum and minimum output.
hi, in the paper it already gave me a sketch but it wasn't really useful since it had no numbers... 0
6 years ago
#4
(Original post by Miss_Jazz)
hi, in the paper it already gave me a sketch but it wasn't really useful since it had no numbers... Well the sketch should still give some big clues to the range. If f(x) move up the y-axis to infinity and down the negative y-axis to infinity then you have a range of ,

the range is where ever the y-axis goes to basically. So look at the graph and see if there is a maximum point/turning point or a minimum.

Edit: just saw you posted the graph, I don't want to give away the answer though. But maybe use a calculator with some really high values of x and see if f(x) is still increasing, if it is, then the range is probably all the real numbers
0
6 years ago
#5
(Original post by Miss_Jazz)
hi, in the paper it already gave me a sketch but it wasn't really useful since it had no numbers...  0
Thread starter 6 years ago
#6
(Original post by hello calum)
Well the sketch should still give some big clues to the range. If f(x) move up the y-axis to infinity and down the negative y-axis to infinity then you have a range of ,

the range is where ever the y-axis goes to basically. So look at the graph and see if there is a maximum point/turning point or a minimum.

Edit: just saw you posted the graph, I don't want to give away the answer though. But maybe use a calculator with some really high values of x and see if f(x) is still increasing, if it is, then the range is probably all the real numbers
in the question it already says that x is equal (that weird E sign) to the real numbers and it wants the answer in terms of pi. I dont want the answers, is there some kind of method to working this out?
0
6 years ago
#7
(Original post by Miss_Jazz)
in the question it already says that x is equal (that weird E sign) to the real numbers and it wants the answer in terms of pi. I dont want the answers, is there some kind of method to working this out?
Well the tan graph has asymptotes at pi/2 and -pi/2 (vertical asymptotes). When drawing the arctan graph the maximum value is pi/2 and the minimum value is -pi/2. When drawing the arcsin, arccos, arctan graphs the stuff on the y axis and the stuff on the x axis is switched in effect.

Just as how the tan graph has f(x) E real numbers, the arctan graph has x E real numbers.

http://www.regentsprep.org/Regents/m...nversetrig.htm

That's how I remember it anyway...

Method is usually draw the graph out and work out the range that way or use the data given in the question i.e. given the domain of the inverse. If you need work out the range or domain of the trigonometric graphs you need to know the sin, cos, tan graphs. From there you can find the domain and range of the inverse functions by using what I said at the beginning (switching x and y thingy)...or you could find the turning points by differentiation.
0
6 years ago
#8
(Original post by Miss_Jazz)
in the question it already says that x is equal (that weird E sign) to the real numbers and it wants the answer in terms of pi. I dont want the answers, is there some kind of method to working this out?
to work it out, you just have to think about what the mazimum and minimum values are, either by differentiating to find turning points, completing the square with a quadratic, or by testing the ends of the domain to see if they have the largest value (or smallest).
0
Thread starter 6 years ago
#9
(Original post by Kishan91)
Well the tan graph has asymptotes at pi/2 and -pi/2 (vertical asymptotes). When drawing the arctan graph the maximum value is pi/2 and the minimum value is -pi/2. When drawing the arcsin, arccos, arctan graphs the stuff on the y axis and the stuff on the x axis is switched in effect.

Just as how the tan graph has f(x) E real numbers, the arctan graph has x E real numbers.

http://www.regentsprep.org/Regents/m...nversetrig.htm

That's how I remember it anyway...

Method is usually draw the graph out and work out the range that way or use the data given in the question i.e. given the domain of the inverse. If you need work out the range or domain of the trigonometric graphs you need to know the sin, cos, tan graphs. From there you can find the domain and range of the inverse functions by using what I said at the beginning (switching x and y thingy)...or you could find the turning points by differentiation.

(Original post by hello calum)
to work it out, you just have to think about what the mazimum and minimum values are, either by differentiating to find turning points, completing the square with a quadratic, or by testing the ends of the domain to see if they have the largest value (or smallest).
Ok, i shall give it a go and see what comes up. Thanks for your help 0
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