g(x)=(x^3)-(3x^2)+17

a)find and classify all critical points of g(x) using exact with x and y values.

b)for what values of x is the function concave down; again exact x and y values.

To find and classify the critical points of the function g(x)=(x^3)-(3x^2)+17, we need to find the points where the derivative of g(x) equals zero or is undefined. Here's how you can do it:

a) Critical Points:

Step 1: Find the derivative of g(x) with respect to x.
g'(x) = 3x^2 - 6x

Step 2: Set g'(x) = 0 and solve for x.
3x^2 - 6x = 0

Step 3: Factor out common terms and solve for x.
3x(x - 2) = 0
=> x = 0 or x - 2 = 0
=> x = 0 or x = 2

The critical points of g(x) are x = 0 and x = 2.

Step 4: Find the corresponding y-values.
Plug the critical points back into the original function g(x) to find the y-values.
For x = 0:
g(0) = (0^3) - (3(0^2)) + 17
= 0 - 0 + 17
= 17

For x = 2:
g(2) = (2^3) - (3(2^2)) + 17
= 8 - 12 + 17
= 13

The critical points with their corresponding x and y-values are:
(0, 17) and (2, 13)

Step 5: Classify the critical points.

To classify the critical points, we need to analyze the behavior of the function around those points. We can do this by examining the sign of the derivative g'(x) in the intervals between the critical points.

• For x < 0: Choose a value of x less than 0, e.g., x = -1.
Substitute this value into g'(x):
g'(-1) = 3(-1)^2 - 6(-1)
= 3(1) + 6
= 3 + 6
= 9

Since g'(-1) > 0, the function is increasing to the left of x = 0. So, x = 0 is a local minimum.

• For 0 < x < 2: Choose a value of x between 0 and 2, e.g., x = 1.
Substitute this value into g'(x):
g'(1) = 3(1)^2 - 6(1)
= 3 - 6
= -3

Since g'(1) < 0, the function is decreasing between x = 0 and x = 2. So, x = 2 is a local maximum.

• For x > 2: Choose a value of x greater than 2, e.g., x = 3.
Substitute this value into g'(x):
g'(3) = 3(3)^2 - 6(3)
= 27 - 18
= 9

Since g'(3) > 0, the function is increasing to the right of x = 2. So, x = 2 is a local minimum.

In summary, the critical points of g(x) are:
Local minimum at (0, 17)
Local maximum at (2, 13)

b) Determining where the function is concave down:

To find where the function g(x) is concave down, we need to determine the intervals where the second derivative is negative. Follow these steps:

Step 1: Find the second derivative of g(x).
g''(x) = 6x - 6

Step 2: Set g''(x) < 0 and solve for x.
6x - 6 < 0

Step 3: Solve the inequality for x.
6x < 6
x < 1

The function g(x) is concave down for x < 1.

In terms of exact x and y values, the function g(x) is concave down for all x-values less than 1.