Respuesta :
SO what you need to do is:
Start with |f(x) - 3| < 0.4
and plug in f(x) = x+1
to get |f(x) – 3| < 0.4
|x+1 – 3| < 0.4
|x - 2| < 0.4
-0.4 < x - 2 < 0.4
-0.4+2 < x < 0.4+2
1.6 < x < 2.4
So delta would be 2.3
Hope this is what you were looking for
Start with |f(x) - 3| < 0.4
and plug in f(x) = x+1
to get |f(x) – 3| < 0.4
|x+1 – 3| < 0.4
|x - 2| < 0.4
-0.4 < x - 2 < 0.4
-0.4+2 < x < 0.4+2
1.6 < x < 2.4
So delta would be 2.3
Hope this is what you were looking for
The largest δ such that if 0 < |x – 2| < δ is 0.4
Further explanation
Solving linear equation mean calculating the unknown variable from the equation.
Let the linear equation : y = mx + c
If we draw the above equation on Cartesian Coordinates , it will be a straight line with :
m → gradient of the line
( 0 , c ) → y - intercept
Gradient of the line could also be calculated from two arbitrary points on line ( x₁ , y₁ ) and ( x₂ , y₂ ) with the formula :
[tex]\large {m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
If point ( x₁ , y₁ ) is on the line with gradient m , the equation of the line will be :
[tex]y - y_1 = m ( x - x_1 )[/tex]
Let us tackle the problem.
Given:
[tex]f(x) = x + 1[/tex]
If [tex]|f(x) - 3| < 0.4[/tex] , then :
[tex]|f(x) - 3| < 0.4[/tex]
[tex]|x + 1 - 3| < 0.4[/tex]
[tex]|x -2| < 0.4[/tex]
Since the absolute value of a function is always positive, then:
[tex]0 < |x -2| < 0.4[/tex]
From the relationship above, then the largest δ such that :
[tex]0 < |x - 2| < \delta[/tex]
→ δ = 0.4
Learn more
- Infinite Number of Solutions : https://brainly.com/question/5450548
- System of Equations : https://brainly.com/question/1995493
- System of Linear equations : https://brainly.com/question/3291576
Answer details
Grade: High School
Subject: Mathematics
Chapter: Linear Equations
Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point
