Find any critical numbers of the function. (Enter your answers as a comma-separated list.)

by critical numbers, they mean

the turning points, (maximim and minimum points)
any points of inflection
the x and the y -intercepts

so for y-intercept, let x = 0
so g(0) = 0, the origin (0,0) is both an x and a y intercept.
for more x-intercepts , let y = 0
x^7 - 7x^5 = 0
x^5(x^2 - 7) = 0
so x = 0 , x = ± √7

g'(x) = 7x^6 - 35x^4 = 0 for max/mins
7x^4(x^2 - 5) = 0
x = 0 or x = ± √5

g''(x) = 42x^5 - 140x^3
= 0 for any points of inflection
14x^3(3x^2 - 10) = 0
x = 0 or x = ± √(10/3)

since (0,0) is both a turning point and a point of inflection, It is neither a maximim nor a minimum

so you have 4 other different x values to sub back into the original equation to find the corresponding y value of the points,
I will leave that up to you

Thank you I really appreciate your help :)

To find the critical numbers of a function, you need to follow these steps:

1. Calculate the derivative of the function.
2. Set the derivative equal to zero and solve for x.
3. The values of x obtained in step 2 are the critical numbers.

Let's work through the steps for the given function g(x) = x^7 − 7x^5.

Step 1: Calculate the derivative of g(x).
To find the derivative of g(x), you can use the power rule, which states that the derivative of x^n is n * x^(n-1).

Taking the derivative of g(x) = x^7 − 7x^5:
g'(x) = 7x^(7-1) - 5(7)x^(5-1)
= 7x^6 - 35x^4

Step 2: Set the derivative equal to zero and solve for x.
Setting g'(x) = 0:
7x^6 - 35x^4 = 0

To solve this equation, you can factor out the common factor x^4:
x^4(7x^2 - 35) = 0

Now, we have two possibilities:
1. x^4 = 0 => x = 0
2. 7x^2 - 35 = 0 => 7x^2 = 35 => x^2 = 5 => x = ±√5

Step 3: Determine the critical numbers.
The critical numbers are the values of x obtained in step 2, which are:
- x = 0
- x = √5
- x = -√5

So, the critical numbers of the given function g(x) = x^7 − 7x^5 are 0, √5, and -√5.

Please note that these are potential critical numbers. To confirm if they are indeed critical points, you should check if the second derivative is positive or negative at each point.