if f(x)= 4x^3-21x^2+36x-4, then the graph of f is decreasing and concave up on the interval(?,?)

i got (3/2,2) but wasn't sure

f = 4x^3-21x^2+36x-4

f' = 12x^2 - 42x + 36 = 6(2x^2-7x+6) = 6(2x-3)(x-2)
f" = 24x-42 = 6(4x-7)

f is concave up for x > 7/4
f is decreasing for x in (3/2,2)

So, both for x in (7/4,2)

f has a max at x=3/2, so it cannot be concave up there.

ok ty!

To determine the intervals on which the graph of a function is decreasing and concave up, we need to analyze the first and second derivatives of the function.

First, let's find the first derivative of f(x). The first derivative gives us information about the slope of the function:

f'(x) = 12x^2 - 42x + 36

Now, to determine where the function is decreasing, we need to find the critical points. The critical points occur where the derivative is equal to zero or undefined.

To find the critical points, we set f'(x) = 0 and solve for x:

12x^2 - 42x + 36 = 0

Using the quadratic formula, x = [-b ± √(b^2 - 4ac)] / (2a), where here a = 12, b = -42, and c = 36:

x = [42 ± √((-42)^2 - 4(12)(36))] / (2 * 12)
x = [42 ± √(1764 - 1728)] / 24
x = [42 ± √36] / 24
x = (42 ± 6) / 24

This gives us two potential critical points: x = 2/3 and x = 3/2.

Next, let's find the second derivative of f(x). The second derivative gives us information about the concavity of the function:

f''(x) = 24x - 42

To determine where the function is concave up, we need to find where f''(x) > 0.

Setting f''(x) > 0 and solving for x:

24x - 42 > 0
24x > 42
x > 42/24
x > 7/4

Now, let's summarize our findings:

1. The critical points we found are x = 2/3 and x = 3/2.
2. The function is concave up when x > 7/4.

Therefore, we can conclude that the graph of f(x) is decreasing and concave up on the interval (7/4, +∞).