f(x) = \frac{ x^3 }{ x^2 - 25 }

defined on the interval [ -18, 18 ].
Enter points, such as inflection points in ascending order, i.e. smallest x values first. Enter intervals in ascending order also.

The function f(x) has vertical asympototes at (? )and (?) .

f(x) is concave up on the region (?)to (?)and (?) to(?) .

The inflection point for this function is (?)

actually i solved my vertical asymptote it's -5 and 5. but how will i find the concave up and inflection point

recall your second derivative

concave up if f" > 0
inflection when f" = 0
concave down if f" < 0

So, you have

f"(x) = 50x(x^2+75)/(x^2-25)^3
f" < 0 on (-∞,-5)
f" > 0 on (-5,0)
f"(0) = 0
f" < 0 on (0,5)
f" > 0 on (5,∞)

So, match that up with the graph of f(x):

http://www.wolframalpha.com/input/?i=x^3%2F%28x^2-25%29

To find the vertical asymptotes of the function f(x), you need to determine the values of x for which the denominator of the function becomes zero. In this case, the denominator is x^2 - 25. Set this equal to zero and solve for x:

x^2 - 25 = 0

This equation can be factored as (x - 5)(x + 5) = 0. So, the vertical asymptotes occur when x = 5 and x = -5.

Next, to determine the regions where f(x) is concave up, you need to find the intervals where the second derivative of the function is positive. The second derivative of f(x) can be found by differentiating the function twice. Let's start with finding the first derivative:

f'(x) = (x^2 - 25)(3x) - (x^3)(2x)
= 3x(x^2 - 25) - 2x^4

Now, differentiate the first derivative to find the second derivative:

f''(x) = 3(x^2 - 25) + 3x(2x) - 8x^3
= 3x^2 - 75 + 6x^2 - 8x^3
= -8x^3 + 9x^2 - 75

To find when the second derivative is positive, solve the inequality:

-8x^3 + 9x^2 - 75 > 0

This is a cubic inequality that can be solved by factoring or using a sign chart. However, factoring or determining the exact intervals where the inequality is true can be quite complex. Instead, you can use a graphing tool or a calculator to visualize the concave up regions. Alternatively, you can use numerical methods to find approximate values of the regions.

Regarding the inflection point, an inflection point occurs when the concavity of a function changes. This happens when the second derivative changes sign. Unfortunately, finding the exact coordinates of an inflection point requires solving a cubic equation, which is not straightforward. Again, using a graphing tool or a calculator can help you approximate the location of the inflection point.

In summary:

- The vertical asymptotes of f(x) are x = 5 and x = -5.
- The regions of f(x) that are concave up cannot be determined explicitly without further analysis.
- The inflection point of the function cannot be determined exactly without solving a cubic equation.