1. Periksa kekontinuan fungsi berikut. Jika kontinu , gunakan definisi kekontinuan dan kriteria barisan untuk membuktikan. (a) f(x)=2x^3+6 pada R (b) g(y)=(2y+5)/(y-16) di y=17 (c) h(z)=(z^2+1)/(z-1) di z=-1 dan z=1

Let's examine the continuity of the given functions.

(a) f(x) = 2x³ + 6 on ℝ

This function is a polynomial. Polynomials are continuous everywhere. We can prove this using the epsilon-delta definition of continuity.

Proof using the epsilon-delta definition:

Let c be any real number. We want to show that for any ε > 0, there exists a δ > 0 such that if 0 < |x - c| < δ, then |f(x) - f(c)| < ε.

|f(x) - f(c)| = |2x³ + 6 - (2c³ + 6)| = |2x³ - 2c³| = 2|x³ - c³| = 2|x - c||x² + xc + c²|

Let's assume |x - c| < 1. Then |x| < |c| + 1. Therefore,

|x² + xc + c²| ≤ |x|² + |x||c| + |c|² < (|c| + 1)² + (|c| + 1)|c| + |c|² = 3|c|² + 3|c| + 1 = M (a constant depending on c)

So, |f(x) - f(c)| < 2M|x - c|.

Now, choose δ = min(1, ε/(2M)). If 0 < |x - c| < δ, then

|f(x) - f(c)| < 2M|x - c| < 2M(ε/(2M)) = ε

This proves that f(x) is continuous at any c ∈ ℝ. Since c was arbitrary, f(x) is continuous on ℝ.


Proof using sequential criterion:

Let {xₙ} be a sequence such that lim (n→∞) xₙ = c, where c ∈ ℝ. Then

lim (n→∞) f(xₙ) = lim (n→∞) (2xₙ³ + 6) = 2(lim (n→∞) xₙ)³ + 6 = 2c³ + 6 = f(c)

Since the limit of the function values along any convergent sequence equals the function value at the limit of the sequence, the function is continuous on ℝ.


(b) g(y) = (2y + 5)/(y - 16) at y = 17

The function g(y) is a rational function. Rational functions are continuous everywhere except where the denominator is zero. The denominator is zero at y = 16. Since we are checking continuity at y = 17, the function is continuous at this point.

Proof using the epsilon-delta definition (sketch): Similar to part (a), but we need to restrict δ to ensure the denominator doesn't approach zero. This is possible since we're evaluating at y=17, which is far from y=16.

Proof using sequential criterion: If {yₙ} is a sequence converging to 17, then the limit of g(yₙ) will be g(17) because the denominator will not approach zero.


(c) h(z) = (z² + 1)/(z - 1) at z = -1 and z = 1

This is another rational function.

* At z = -1: The denominator is -2, which is non-zero. Therefore, h(z) is continuous at z = -1. The proof using either the epsilon-delta definition or the sequential criterion is similar to part (b).

* At z = 1: The denominator is zero. Therefore, h(z) is *not* continuous at z = 1. The limit as z approaches 1 does not exist (it approaches ±∞). The function is undefined at z=1.


In summary:

* f(x) is continuous on ℝ.
* g(y) is continuous at y = 17.
* h(z) is continuous at z = -1 but not continuous at z = 1.