Complete Question
The complete question is shown on the first uploaded image
Answer:
Interval Slope of secant lines
[1,2] 113.000
[1, 1.5] 75.000
[1, 1.1 ] 51.960
[1, 1.01] 47.482
[1, 1.001] 47.048
Step-by-step explanation:
From the question we are told that
The function is [tex]f(x) = 16x^3 -x[/tex]
Now considering the intervals
For [1, 2]
The slopes of secant lines is mathematically represented as
[tex]\frac{ f(2) - f(1)}{2-1} = \frac{16(2)^3 - 2 - [16(1)^3 -1] }{2-1 } = 113.000[/tex]
For [1, 1.5]
The slopes of secant lines is mathematically represented as
[tex]\frac{ f(1.5) - f(1)}{1.5-1} = \frac{16(1.5)^3 - 1.5 - [16(1)^3 -1] }{1.5-1 } = 75.000[/tex]
For [1, 1.1]
The slopes of secant lines is mathematically represented as
[tex]\frac{ f(1.1) - f(1)}{1.1-1} = \frac{16(1.1)^3 - 1.1 - [16(1)^3 -1] }{1.1-1 } = 51.960[/tex]
For [1, 1.01]
The slopes of secant lines is mathematically represented as
[tex]\frac{ f(1.01) - f(1)}{1.01-1} = \frac{16(1.01)^3 - 1.01 - [16(1)^3 -1] }{1.01-1 } =47.482[/tex]
For [1, 1.001]
The slopes of secant lines is mathematically represented as
[tex]\frac{ f(1.001) - f(1)}{1.001-1} = \frac{16(1.01)^3 - 1.001 - [16(1)^3 -1] }{1.001-1 } =47.048[/tex]
