Consider the function \( f(x) = 12x^5 + 45x^4 - 80x^3 + 1 \).

For this function, there are four important intervals: \((-∞, A]\), \([A, B]\), \([B, C]\), and \([C, ∞)\), where \(A\), \(B\), and \(C\) are the critical numbers.

1. Find \(A\), \(B\), and \(C\).
2. At each critical number \(A\), \(B\), and \(C\), does \(f(x)\) have a local minimum, a local maximum, or neither?

Type in your answer as LMIN, LMAX, or NEITHER.

- At \(A\):
- At \(B\):
- At \(C\):

Consider the function, f(x) = 12x5 + 45x4 - 80x³ + 1. For this function, there are four important intervals: (-[infinity], A], [A, B], [B, C], and [C, [infinity]) where A, B, and C are the critical numbers.

Find A, B, and C: Now we will differentiate the given function, f(x). So, f′(x) = 60x4 + 180x³ - 240x²

Now we will equate this to zero. 60x4 + 180x³ - 240x² = 0Factor out the GCF 60x², 60x² (x² + 3x - 4) = 0 => 60x²(x + 4) (x - 1) = 0

So, the critical points are x = -4, 0 and 1. Hence, A, B, and C are A = -4, B = 0, and C = 1.

Now, we will find whether f(x) has a local max or min or neither at A, B, and C.At A: f(-4) = 9673 ⇒ LMINAt B: f(0) = 1 ⇒ NEITHERAt C: f(1) = -22 ⇒ LMAX Therefore, At A, f(x) has a local min.

At B, f(x) has neither local max nor local min. At C, f(x) has a local max.

So, the answer is, At A: LMINAt B: NEITHERAt C: LMAX

To know more about intervals visit:

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