# How do you solve the inequality #9x^2-6x+1<=0#?

Actually the left side can never be less than 0 for real numbers. It's lowest value is

You can see that from a diagram:

Since this is precalculus, I'm in doubt if derivation should be used in the solution, but using it you can show that a tangent at

Other than that we can write:

Since the left hand of the inequality is a square, we can conclude that it will never be negative, and it's lowest value is

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To solve the inequality (9x^2 - 6x + 1 \leq 0), you can follow these steps:

- Find the roots of the quadratic equation (9x^2 - 6x + 1 = 0). You can use the quadratic formula or factoring.
- Once you have the roots, use them to determine the intervals where the quadratic expression is less than or equal to zero.
- Test a value within each interval to determine whether the expression is positive or negative.
- Identify the intervals where the expression is less than or equal to zero.

Then, you'll have the solution to the inequality in terms of intervals on the number line.

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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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