Find the Critical Numbers?

1.) f(x) = 2x^3 + 3x^2 + 36x
x=?
2.) g(t) = |5t − 4|
t=?

No ideas at all?

Critical numbers are min, max, point of inflection, roots

min and max occur where f' = 0
inflection is where f'' = 0 and f' ≠ zero

The absolute value problem is tricky, since you can't rely on f' to get the min/max.

Observe, however, that |5t-4| is always at least zero.

It is zero only where 5t-4 = 0.

To find the critical numbers for each function, we need to first find the derivative of each function and set them equal to zero. The critical numbers will be the values of x or t that satisfy this equation.

1.) For f(x) = 2x^3 + 3x^2 + 36x, let's find its derivative, f'(x):

f'(x) = 6x^2 + 6x + 36

Next, to find the critical numbers, set f'(x) equal to zero and solve for x:

6x^2 + 6x + 36 = 0

Now we can solve this quadratic equation. We can either use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values for a, b, and c:

x = (-6 ± √(6^2 - 4*6*36)) / (2*6)

Simplifying:

x = (-6 ± √(36 - 864)) / 12
x = (-6 ± √(-828)) / 12

Since the discriminant (√(b^2 - 4ac)) is negative, it means that there are no real solutions for x. Therefore, there are no critical numbers for f(x) = 2x^3 + 3x^2 + 36x.

2.) For g(t) = |5t − 4|, let's find its derivative, g'(t):

Since g(t) contains an absolute value function, we will need to consider two cases - when (5t - 4) is positive and when it is negative.

Case 1: When (5t - 4) > 0, the derivative is simply:

g'(t) = 5

Case 2: When (5t - 4) < 0, we need to remember that the absolute value of a negative number is positive, so:

g'(t) = -5

Now we can set g'(t) equal to zero and solve for t:

Case 1: 5 = 0 (No solution)
Case 2: -5 = 0 (No solution)

Since there are no values of t that make g(t) = |5t − 4| have zero slope, there are no critical numbers for g(t).