How do you find the critical points for the inequality #(2x+1)/(x-9)>=0#?

Answer 1

See the explanation.

I think the question wants the key numbers or the partition numbers.

These are the values of #x# at which the sign of the expression MIGHT change.
The sign of a rational function MIGHT change when either the numerator is #0# or when the denominator is #0# (where the expression is not defined.)
For the expression #(2x+1)/(x-9)#,
the sign MIGHT and does change at #x=-1/2# (the solution to #2x+1=0# and
at #x=9# (the solution to #x-9=0#).
In the graph of #y=(2x+1)/(x-9)# below, you can see where the expression changes sign:

graph{y=(2x+1)/(x-9) [-25.9, 39.05, -23.36, 9.1]}

To the left of #x=-1/2#, y is positive. Then at #x=-1/2#, #y# changes from positive to negative. (You can use a mouse to scroll in and to drag the graph around.) #y# stays negative until we get to #x=9# where #y# suddenly becomes very,very positive.
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Answer 2
To find the critical points for the inequality \(\frac{2x+1}{x-9} \geq 0\), follow these steps: 1. Determine where the expression \(\frac{2x+1}{x-9}\) equals zero or undefined. This will give you critical points. 2. Set the numerator, \(2x+1\), equal to zero and solve for \(x\). \(2x + 1 = 0\) \(2x = -1\) \(x = -\frac{1}{2}\) 3. Set the denominator, \(x - 9\), equal to zero and solve for \(x\). \(x - 9 = 0\) \(x = 9\) So, the critical points are \(x = -\frac{1}{2}\) and \(x = 9\). 4. Test intervals between and outside of the critical points by choosing test points within those intervals and evaluating the expression \(\frac{2x+1}{x-9}\) to determine the sign. - Choose a test point \(x < -\frac{1}{2}\), for example, \(x = -1\). - Substitute \(x = -1\) into \(\frac{2x+1}{x-9}\): \(\frac{2(-1)+1}{(-1)-9} = \frac{-1}{-10} = \frac{1}{10}\), which is positive. - Choose a test point \(-\frac{1}{2} < x < 9\), for example, \(x = 0\). - Substitute \(x = 0\) into \(\frac{2x+1}{x-9}\): \(\frac{2(0)+1}{(0)-9} = \frac{1}{-9}\), which is negative. - Choose a test point \(x > 9\), for example, \(x = 10\). - Substitute \(x = 10\) into \(\frac{2x+1}{x-9}\): \(\frac{2(10)+1}{(10)-9} = \frac{21}{1}\), which is positive. 5. Based on the signs obtained in step 4, determine the intervals where the inequality is true. - The inequality is true for \(x < -\frac{1}{2}\) and \(x > 9\). So, the solution to the inequality \(\frac{2x+1}{x-9} \geq 0\) is \(x \in (-\infty, -\frac{1}{2}] \cup (9, \infty)\).
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Answer 3

To find the critical points for the inequality (\frac{{2x+1}}{{x-9}} \geq 0), we first need to determine where the expression is equal to zero and where it is undefined.

Setting the numerator equal to zero: [2x + 1 = 0] [2x = -1] [x = -\frac{1}{2}]

Setting the denominator equal to zero: [x - 9 = 0] [x = 9]

Now, we have two critical points: (x = -\frac{1}{2}) and (x = 9). We need to test the intervals between and beyond these critical points to determine where the expression is positive or negative.

Testing the interval (x < -\frac{1}{2}): Choose a test point (x = -1). Substitute it into the inequality: [\frac{2(-1)+1}{(-1)-9} = \frac{-1}{-10} > 0] Since the expression is positive in this interval, it satisfies the inequality.

Testing the interval (-\frac{1}{2} < x < 9): Choose a test point (x = 0). Substitute it into the inequality: [\frac{2(0)+1}{(0)-9} = \frac{1}{-9} < 0] Since the expression is negative in this interval, it does not satisfy the inequality.

Testing the interval (x > 9): Choose a test point (x = 10). Substitute it into the inequality: [\frac{2(10)+1}{(10)-9} = \frac{21}{1} > 0] Since the expression is positive in this interval, it satisfies the inequality.

Therefore, the critical points where the inequality holds true are (x \leq -\frac{1}{2}) and (x \geq 9).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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