For the function [tex]f(x) = 4x^2 + 7x - 5[/tex]:

a) Find the slope of the secant line from [tex]x = 7[/tex] to [tex]x = 11[/tex].

b) Create a table showing the slopes of the secant lines to [tex]f(x)[/tex] between 3 and [tex]x[/tex] as [tex]x[/tex] approaches 3 from the right. (Include enough decimal places in your answers.)

c) Explain how you can use the table you created in part b) to find the slope of the tangent to [tex]f(x)[/tex] at [tex]x = 3[/tex]. State the slope of the tangent at [tex]x = 3[/tex].



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].