Yes, 23 has an inverse mod 1000 because gcd(23, 1000) = 1 (i.e. they are coprime).
Let <em>x</em> be the inverse. Then <em>x</em> is such that
23<em>x</em> ≡ 1 (mod 1000)
Use the Euclidean algorithm to solve for <em>x</em> :
1000 = 43×23 + 11
23 = 2×11 + 1
→ 1 ≡ 23 - 2×11 (mod 1000)
→ 1 ≡ 23 - 2×(1000 - 43×23) (mod 1000)
→ 1 ≡ 23 - 2×1000 + 86×23 (mod 1000)
→ 1 ≡ 87×23 - 2×1000 ≡ 87×23 (mod 1000)
→ 23⁻¹ ≡ 87 (mod 1000)
Answer:
- (-1, -32) absolute minimum
- (0, 0) relative maximum
- (2, -32) absolute minimum
- (+∞, +∞) absolute maximum (or "no absolute maximum")
Step-by-step explanation:
There will be extremes at the ends of the domain interval, and at turning points where the first derivative is zero.
The derivative is ...
h'(t) = 24t^2 -48t = 24t(t -2)
This has zeros at t=0 and t=2, so that is where extremes will be located.
We can determine relative and absolute extrema by evaluating the function at the interval ends and at the turning points.
h(-1) = 8(-1)²(-1-3) = -32
h(0) = 8(0)(0-3) = 0
h(2) = 8(2²)(2 -3) = -32
h(∞) = 8(∞)³ = ∞
The absolute minimum is -32, found at t=-1 and at t=2. The absolute maximum is ∞, found at t→∞. The relative maximum is 0, found at t=0.
The extrema are ...
- (-1, -32) absolute minimum
- (0, 0) relative maximum
- (2, -32) absolute minimum
- (+∞, +∞) absolute maximum
_____
Normally, we would not list (∞, ∞) as being an absolute maximum, because it is not a specific value at a specific point. Rather, we might say there is no absolute maximum.