Use the given graph of f to find a number δ such that
if
0 < |x − 3| < δ then |f(x) − 2| < 0.5.
since i cannot put up the graph ill put the coordinates
(2.6,1.5)
(3,2)
(3.8,2.5)
Please help im not sure how to solve this
To find a suitable value for δ, we need to analyze the graph of f and determine the behavior of the function around the point (3, 2).
Based on the given coordinates, we can observe that when x is close to 3, the corresponding values of f(x) are close to 2. This suggests that the function f is relatively stable and does not vary significantly around x = 3.
To ensure that |f(x) - 2| is less than 0.5 when 0 < |x - 3| < δ, we need to find a value of δ such that as x approaches 3, f(x) remains within the range (1.5, 2.5).
Looking at the coordinates (2.6, 1.5) and (3.8, 2.5), we can determine that the function f(x) has a maximum value of 2.5 and a minimum value of 1.5 within the interval [2.6, 3.8].
Since we want |f(x) - 2| to be less than 0.5, we can set a maximum deviation of 0.5 around the value of 2. Therefore, we can choose δ as the minimum distance between x = 3 and the nearest point where |f(x) - 2| is equal to or greater than 0.5.
In this case, the nearest points are (2.6, 1.5) and (3.8, 2.5). So, the minimum distance is given by:
δ = min(|3 - 2.6|, |3 - 3.8|) = min(0.4, 0.8) = 0.4
Hence, we can choose δ = 0.4 as the number that satisfies the condition:
if 0 < |x - 3| < 0.4, then |f(x) - 2| < 0.5.
Please note that the graph coordinates provided are crucial for this analysis.