Answer :
Sure! Let's go through each part of the question step by step.
### Part (a): Slope of the Secant Line
To find the slope of the secant line for the function [tex]\( f(x) = 4x^2 + 7x - 5 \)[/tex] between [tex]\( x=7 \)[/tex] and [tex]\( x=11 \)[/tex]:
1. Evaluate the function at [tex]\( x=7 \)[/tex]:
[tex]\[ f(7) = 4(7)^2 + 7(7) - 5 = 4 \cdot 49 + 49 - 5 = 196 + 49 - 5 = 240 \][/tex]
2. Evaluate the function at [tex]\( x=11 \)[/tex]:
[tex]\[ f(11) = 4(11)^2 + 7(11) - 5 = 4 \cdot 121 + 77 - 5 = 484 + 77 - 5 = 556 \][/tex]
3. Calculate the slope of the secant line using the formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[ \text{slope} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} = \frac{f(11) - f(7)}{11 - 7} = \frac{556 - 240}{11 - 7} = \frac{316}{4} = 79.0 \][/tex]
So, the slope of the secant line from [tex]\( x=7 \)[/tex] to [tex]\( x=11 \)[/tex] is [tex]\( 79.0 \)[/tex].
### Part (b): Table of Secant Slopes Approaching [tex]\( x=3 \)[/tex] from the Right
To create a table showing the slopes as [tex]\( x \)[/tex] approaches 3 from the right, we consider [tex]\( x \)[/tex] values incrementing by 0.1 starting from just greater than 3 (i.e., [tex]\( 3.1 \)[/tex]) up to [tex]\( 5.0 \)[/tex].
Here are the slopes of the secant lines between [tex]\( x=3 \)[/tex] and [tex]\( x\)[/tex] where [tex]\( x \)[/tex] is incrementing by 0.1:
- At [tex]\( x = 3.1 \)[/tex], slope = 31.40
- At [tex]\( x = 3.2 \)[/tex], slope = 31.80
- At [tex]\( x = 3.3 \)[/tex], slope = 32.20
- At [tex]\( x = 3.4 \)[/tex], slope = 32.60
- At [tex]\( x = 3.5 \)[/tex], slope = 33.00
- At [tex]\( x = 3.6 \)[/tex], slope = 33.40
- At [tex]\( x = 3.7 \)[/tex], slope = 33.80
- At [tex]\( x = 3.8 \)[/tex], slope = 34.20
- At [tex]\( x = 3.9 \)[/tex], slope = 34.60
- At [tex]\( x = 4.0 \)[/tex], slope = 35.00
- At [tex]\( x = 4.1 \)[/tex], slope = 35.40
- At [tex]\( x = 4.2 \)[/tex], slope = 35.80
- At [tex]\( x = 4.3 \)[/tex], slope = 36.20
- At [tex]\( x = 4.4 \)[/tex], slope = 36.60
- At [tex]\( x = 4.5 \)[/tex], slope = 37.00
- At [tex]\( x = 4.6 \)[/tex], slope = 37.40
- At [tex]\( x = 4.7 \)[/tex], slope = 37.80
- At [tex]\( x = 4.8 \)[/tex], slope = 38.20
- At [tex]\( x = 4.9 \)[/tex], slope = 38.60
- At [tex]\( x = 5.0 \)[/tex], slope = 39.00
### Part (c): Using the Table to Find the Slope of the Tangent at [tex]\( x=3 \)[/tex]
By observing the table created in part (b), we can see how the slopes of the secant lines change as [tex]\( x \)[/tex] approaches 3 from the right. As [tex]\( x \)[/tex] gets closer to 3, the slopes gradually approach a single value.
#### Conclusion on the Tangent Slope at [tex]\( x=3 \)[/tex]:
To find the exact slope of the tangent line at [tex]\( x=3 \)[/tex], you notice that these slopes appear to be converging to a common value. To confirm, we can analytically determine the derivative of the function at [tex]\( x=3 \)[/tex]:
1. Differentiate the function [tex]\( f(x) = 4x^2 + 7x - 5 \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(4x^2 + 7x - 5) = 8x + 7 \][/tex]
2. Evaluate the derivative at [tex]\( x=3 \)[/tex]:
[tex]\[ f'(3) = 8(3) + 7 = 24 + 7 = 31.0 \][/tex]
Therefore, the slope of the tangent to the function [tex]\( f(x) \)[/tex] at [tex]\( x=3 \)[/tex] is [tex]\( 31.0 \)[/tex].
### Part (a): Slope of the Secant Line
To find the slope of the secant line for the function [tex]\( f(x) = 4x^2 + 7x - 5 \)[/tex] between [tex]\( x=7 \)[/tex] and [tex]\( x=11 \)[/tex]:
1. Evaluate the function at [tex]\( x=7 \)[/tex]:
[tex]\[ f(7) = 4(7)^2 + 7(7) - 5 = 4 \cdot 49 + 49 - 5 = 196 + 49 - 5 = 240 \][/tex]
2. Evaluate the function at [tex]\( x=11 \)[/tex]:
[tex]\[ f(11) = 4(11)^2 + 7(11) - 5 = 4 \cdot 121 + 77 - 5 = 484 + 77 - 5 = 556 \][/tex]
3. Calculate the slope of the secant line using the formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[ \text{slope} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} = \frac{f(11) - f(7)}{11 - 7} = \frac{556 - 240}{11 - 7} = \frac{316}{4} = 79.0 \][/tex]
So, the slope of the secant line from [tex]\( x=7 \)[/tex] to [tex]\( x=11 \)[/tex] is [tex]\( 79.0 \)[/tex].
### Part (b): Table of Secant Slopes Approaching [tex]\( x=3 \)[/tex] from the Right
To create a table showing the slopes as [tex]\( x \)[/tex] approaches 3 from the right, we consider [tex]\( x \)[/tex] values incrementing by 0.1 starting from just greater than 3 (i.e., [tex]\( 3.1 \)[/tex]) up to [tex]\( 5.0 \)[/tex].
Here are the slopes of the secant lines between [tex]\( x=3 \)[/tex] and [tex]\( x\)[/tex] where [tex]\( x \)[/tex] is incrementing by 0.1:
- At [tex]\( x = 3.1 \)[/tex], slope = 31.40
- At [tex]\( x = 3.2 \)[/tex], slope = 31.80
- At [tex]\( x = 3.3 \)[/tex], slope = 32.20
- At [tex]\( x = 3.4 \)[/tex], slope = 32.60
- At [tex]\( x = 3.5 \)[/tex], slope = 33.00
- At [tex]\( x = 3.6 \)[/tex], slope = 33.40
- At [tex]\( x = 3.7 \)[/tex], slope = 33.80
- At [tex]\( x = 3.8 \)[/tex], slope = 34.20
- At [tex]\( x = 3.9 \)[/tex], slope = 34.60
- At [tex]\( x = 4.0 \)[/tex], slope = 35.00
- At [tex]\( x = 4.1 \)[/tex], slope = 35.40
- At [tex]\( x = 4.2 \)[/tex], slope = 35.80
- At [tex]\( x = 4.3 \)[/tex], slope = 36.20
- At [tex]\( x = 4.4 \)[/tex], slope = 36.60
- At [tex]\( x = 4.5 \)[/tex], slope = 37.00
- At [tex]\( x = 4.6 \)[/tex], slope = 37.40
- At [tex]\( x = 4.7 \)[/tex], slope = 37.80
- At [tex]\( x = 4.8 \)[/tex], slope = 38.20
- At [tex]\( x = 4.9 \)[/tex], slope = 38.60
- At [tex]\( x = 5.0 \)[/tex], slope = 39.00
### Part (c): Using the Table to Find the Slope of the Tangent at [tex]\( x=3 \)[/tex]
By observing the table created in part (b), we can see how the slopes of the secant lines change as [tex]\( x \)[/tex] approaches 3 from the right. As [tex]\( x \)[/tex] gets closer to 3, the slopes gradually approach a single value.
#### Conclusion on the Tangent Slope at [tex]\( x=3 \)[/tex]:
To find the exact slope of the tangent line at [tex]\( x=3 \)[/tex], you notice that these slopes appear to be converging to a common value. To confirm, we can analytically determine the derivative of the function at [tex]\( x=3 \)[/tex]:
1. Differentiate the function [tex]\( f(x) = 4x^2 + 7x - 5 \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(4x^2 + 7x - 5) = 8x + 7 \][/tex]
2. Evaluate the derivative at [tex]\( x=3 \)[/tex]:
[tex]\[ f'(3) = 8(3) + 7 = 24 + 7 = 31.0 \][/tex]
Therefore, the slope of the tangent to the function [tex]\( f(x) \)[/tex] at [tex]\( x=3 \)[/tex] is [tex]\( 31.0 \)[/tex].
