Answer:
(a) To graph the function f(x) = x + 1, we need to plot points on a coordinate plane. The given function is a linear function with a slope of 1 and a y-intercept of 1. We can choose different values of x and calculate the corresponding y-values to plot the points.
Let's choose a few values for x:
- If x = -2, then f(x) = (-2) + 1 = -1. So, we have the point (-2, -1).
- If x = -1, then f(x) = (-1) + 1 = 0. So, we have the point (-1, 0).
- If x = 0, then f(x) = 0 + 1 = 1. So, we have the point (0, 1).
- If x = 1, then f(x) = 1 + 1 = 2. So, we have the point (1, 2).
Now, we can plot these points on the coordinate plane and connect them to get the graph of the function f(x) = x + 1.
(b) Now, let's consider the value of a = -1 and the chosen value of L = f(-1) = 2.
We need to show that there is no possible value of d such that if |r - (-1)| < 8, then |f(r) - 2| < d.
By using the hint given, we need to find a point r such that r - 2 < 8 but |f(r) - 2| ≥ d for any value of d.
Let's choose r = 10. In this case, |r - (-1)| = |10 - (-1)| = 11 < 8, which satisfies the condition.
Now, let's calculate |f(r) - 2|:
f(r) = r + 1 = 10 + 1 = 11.
|f(r) - 2| = |11 - 2| = 9 ≥ d.
Since we found a value of r that satisfies the condition |r - (-1)| < 8 but |f(r) - 2| ≥ d, this shows that there is no possible value of d that satisfies the given condition.
(c) From the graph of the function f(x) = x + 1, it appears that as x approaches negative infinity, the value of f(x) approaches a certain value. Let's call this value L.
By observing the graph, a better guess for what L = lim,x→-∞ f(x) should be is L = -∞. This means that as x gets smaller and smaller (approaching negative infinity), f(x) becomes infinitely negative.
To find a value of d such that if |r - (-1)| < d, then |f(r) - L| < 1, we can choose a small value of d, such as d = 0.1.
For example, if r = -1.1, then |r - (-1)| = |-1.1 - (-1)| = 0.1 < d. Also, |f(r) - L| = |(-1.1) + 1 - (-∞)| = |-0.1 - (-∞)| = |-0.1 + ∞| = ∞ < 1.
By choosing a small enough value of d, we can ensure that the condition |r - (-1)| < d will always be satisfied, and the condition |f(r) - L| < 1 will also be satisfied.