Answer:
Step-by-step explanation:
Given function is f(x) = x² - 5x + 5
(a). Slope of the secant of the function in the interval [1, 3],
m = [tex]\frac{\triangle y}{\triangle x}[/tex]
For x = 1
[tex]y_1[/tex] = (1)²- 5(1) + 5
[tex]y_1[/tex] = 1
For x = 3,
[tex]y_2[/tex] = (3)² - 5(3) + 5
[tex]y_2[/tex] = 9 - 15 + 5
[tex]y_2[/tex] = -1
Slope 'm' = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]
m = [tex]\frac{1+1}{1-3}[/tex]
m = -1
(b). Slope of the secant of the function in the interval [1, 2],
m = [tex]\frac{\triangle y}{\triangle x}[/tex]
For x = 1,
[tex]y_1[/tex] = 1 [From part a]
For x = 2,
[tex]y_2=[/tex] (2)² - 5(2) + 5
= 4 - 10 + 5
= -1
Slope = [tex]\frac{-1-1}{2-1}[/tex]
= -2
(c). Slope of the secant of the function in the interval [0, 1]
For x = 0,
[tex]y_1=0-5(0)+5[/tex]
= 5
For x = 1
[tex]y_2=-1[/tex] [From part (b)]
m = [tex]\frac{5+1}{0-1}[/tex]
= -6
(d). Derivative of the function 'f',
f'(x) = 2x - 5
At x = 1,
f'(x) = 2(1) - 5
= -3
Slope of the tangent = -3
a) The slope of the secant line of the function on the interval [1, 3] is -1.
b) The slope of the secant line of the function on the interval [1, 2] is -2.
c) The slope of the secant line of the function on the interval [0, 1] is -4.
d) The estimated slope of the tangent line of [tex]f(x)[/tex] at [tex]x = 1[/tex] is -3.
Let [tex]f(x)[/tex] be a differentiable curve, the slope of a line secant to that curve ([tex]m_{AB}[/tex]) is defined by the following expression:
[tex]m_{AB} = \frac{f(b)-f(a)}{b-a}[/tex], [tex]a \ne b[/tex] (1)
If we know that [tex]f(x) = x^{2}-5\cdot x + 5[/tex], [tex]a = 1[/tex] and [tex]b = 3[/tex], then the slope of the secant line is:
[tex]m_{AB} = \frac{-1-1}{3-1}[/tex]
[tex]m_{AB} = -1[/tex]
The slope of the secant line of the function on the interval [1, 3] is -1. [tex]\blacksquare[/tex]
If we know that [tex]f(x) = x^{2}-5\cdot x + 5[/tex], [tex]a = 1[/tex] and [tex]b = 2[/tex], then the slope of the secant line is:
[tex]m_{AB} = \frac{-1-1}{2-1}[/tex]
[tex]m_{AB} = -2[/tex]
The slope of the secant line of the function on the interval [1, 2] is -2. [tex]\blacksquare[/tex]
If we know that [tex]f(x) = x^{2}-5\cdot x + 5[/tex], [tex]a = 0[/tex] and [tex]b = 1[/tex], then the slope of the secant line is:
[tex]m_{AB} = \frac{1-5}{1-0}[/tex]
[tex]m_{AB} = -4[/tex]
The slope of the secant line of the function on the interval [0, 1] is -4. [tex]\blacksquare[/tex]
The slope of the tangent line of [tex]f(x)[/tex] at [tex]x = 1[/tex] can be estimated by concept of weighted averages, whose application is shown below:
[tex]m = \left(\frac{1-0}{2-0}\right)\cdot (-4) + \left(\frac{2-1}{2-0} \right)\cdot (-2)[/tex]
[tex]m = -3[/tex]
The estimated slope of the tangent line of [tex]f(x)[/tex] at [tex]x = 1[/tex] is -3. [tex]\blacksquare[/tex]
To learn more on secant lines, we kindly invite to check this verified question: https://brainly.com/question/10128640
