Need a bit of help.
Trying to find the image sets of two recipricol functions.
The first g(x)=1/(x-2)+5 (x not equal to 2)
The second g(x)=1/(x-2)+5 (x>2)
Now its quite easy to draw/sketch the graph but ive not really dealt with image sets of recipricol f functions before and so im getting a bit confused where to look.
I know the image set is the y axis and the domain is the x axis so...
am I right if I say
the first image set where x cant equal 2 is this:
[-∞,5)∪[5,∞)
and the second is where I get very lost.
From looking at the graph it seems that the image set is again
[5,∞)
When x > 2.
But it doesnt seem correct to me.
To find the image sets of reciprocal functions, you need to consider the range of possible outputs or y-values for different inputs or x-values.
Let's analyze the first function: g(x) = 1/(x-2) + 5 (x not equal to 2). To find the image set or range of this function, we need to find all possible values that g(x) can take.
When x approaches 2 from the left side (x < 2), the function approaches negative infinity. As x moves further away from 2 towards negative infinity, the function value becomes increasingly negative.
When x approaches 2 from the right side (x > 2), the function approaches positive infinity. As x moves further away from 2 towards positive infinity, the function value becomes increasingly positive.
Therefore, the range of the function g(x) = 1/(x-2) + 5 (x not equal to 2) is (-∞, 5) ∪ (5, +∞). This means that any value less than 5 or greater than 5 can be a possible output of this function, excluding the value 5 itself.
Now let's analyze the second function: g(x) = 1/(x-2) + 5 (x > 2). In this case, the domain is restricted to x values greater than 2.
When x approaches 2 from the right side (x > 2), the function approaches positive infinity. As x moves further away from 2 towards positive infinity, the function value becomes increasingly positive.
Therefore, the range of the function g(x) = 1/(x-2) + 5 (x > 2) is (5, +∞). This means that any value greater than 5 can be a possible output of this function.
In summary:
- The image set or range of the first function g(x) = 1/(x-2) + 5 (x not equal to 2) is (-∞, 5) ∪ (5, +∞).
- The image set or range of the second function g(x) = 1/(x-2) + 5 (x > 2) is (5, +∞).
It seems that you got the range for the first function correct, but there was a misunderstanding regarding the range for the second function.