solve the follow inequality. write your solution in interval notation.

x^3-49x¡Ü0

I think that was supposed to come out as

x^3 - 49x < 0
or
x^3 - 49x < 0
or
x^3 - 49x > 0
or
x^3 - 49x > 0

btw, did I not solve this same question for you on Sunday? Did you not look at it??

http://www.jiskha.com/display.cgi?id=1348975741

it was suppose to be x^3-49x<=0

I didn't know how to make the sign

To solve the inequality x^3 - 49x ≤ 0, we can first find the critical points where the left side of the inequality is equal to zero. This can be done by factoring out x from the equation:

x(x^2 - 49) ≤ 0

Now, notice that x^2 - 49 is a difference of squares and can be factored further as (x - 7)(x + 7). Therefore, the inequality becomes:

x(x - 7)(x + 7) ≤ 0

To find the intervals where the inequality is true, we can create a number line and determine the sign of the expression x(x - 7)(x + 7) for different intervals.

-7 0 7

When x < -7, all three factors x, (x - 7), and (x + 7) are negative, so the product x(x - 7)(x + 7) is negative.

When -7 < x < 0, the factor x is positive, while both (x - 7) and (x + 7) are negative. Therefore, the product x(x - 7)(x + 7) is positive.

When 0 < x < 7, the factors x and (x + 7) are positive, while (x - 7) is negative. Hence, the product x(x - 7)(x + 7) is negative.

When x > 7, all three factors x, (x - 7), and (x + 7) are positive, resulting in a positive product.

To summarize:

-7 < x < 0 or 7 < x

In interval notation, the solution to the inequality x^3 - 49x ≤ 0 is (-7, 0) ∪ (7, ∞).