Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
To solve the given question, let's start with part (a):
Part (a): Find the derivative function [tex]\( f' \)[/tex] for the function [tex]\( f \)[/tex].
The function provided is [tex]\( f(x) = \frac{4}{5x + 1} \)[/tex].
The goal is to find the derivative [tex]\( f'(x) \)[/tex]. The derivative of a function [tex]\( \frac{a}{b(x)} \)[/tex] can be found using the quotient rule or by recognizing it as a constant times the derivative of [tex]\( b(x)^{-1} \)[/tex].
For [tex]\( f(x) = \frac{4}{5x + 1} \)[/tex], we have:
[tex]\[ f(x) = 4 \cdot (5x + 1)^{-1} \][/tex]
Using the chain rule, where [tex]\( g(x) = (5x + 1)^{-1} \)[/tex], we have:
[tex]\[ g(x) = (5x + 1)^{-1} \][/tex]
[tex]\[ g'(x) = -\frac{1}{(5x + 1)^2} \cdot \frac{d}{dx}(5x + 1) \][/tex]
[tex]\[ \frac{d}{dx}(5x + 1) = 5 \][/tex]
So,
[tex]\[ g'(x) = -\frac{5}{(5x + 1)^2} \][/tex]
Thus,
[tex]\[ f'(x) = 4 \cdot g'(x) \][/tex]
[tex]\[ f'(x) = 4 \cdot \left( -\frac{5}{(5x + 1)^2} \right) \][/tex]
[tex]\[ f'(x) = -\frac{20}{(5x + 1)^2} \][/tex]
Therefore,
[tex]\[ f'(x) = -\frac{20}{(5x + 1)^2} \][/tex]
So,
[tex]\[ f'(x) = -\boxed{\frac{20}{(5x + 1)^2}} \][/tex]
Part (b): Determine an equation of the line tangent to the graph of [tex]\( f \)[/tex] at [tex]\( (a, f(a)) \)[/tex] for the given value of [tex]\(a = -1 \)[/tex].
First, we'll find [tex]\( f(a) \)[/tex]. With [tex]\( a = -1 \)[/tex]:
[tex]\[ f(a) = f(-1) = \frac{4}{5(-1) + 1} \][/tex]
[tex]\[ f(-1) = \frac{4}{-5 + 1} \][/tex]
[tex]\[ f(-1) = \frac{4}{-4} \][/tex]
[tex]\[ f(-1) = -1 \][/tex]
We already found the derivative function [tex]\( f'(x) \)[/tex]. Now, we need to compute [tex]\( f'(a) \)[/tex] for [tex]\( a = -1 \)[/tex]:
[tex]\[ f'(-1) = -\frac{20}{(5(-1) + 1)^2} \][/tex]
[tex]\[ f'(-1) = -\frac{20}{(-5 + 1)^2} \][/tex]
[tex]\[ f'(-1) = -\frac{20}{(-4)^2} \][/tex]
[tex]\[ f'(-1) = -\frac{20}{16} \][/tex]
[tex]\[ f'(-1) = -\frac{5}{4} \][/tex]
We now have the slope of the tangent line at [tex]\( x = -1 \)[/tex]:
[tex]\[ m = f'(-1) = -\frac{5}{4} \][/tex]
The equation of the tangent line can be formulated using the point-slope form of the equation of a line:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
With [tex]\( (x_1, y_1) = (-1, -1) \)[/tex]:
[tex]\[ y - (-1) = -\frac{5}{4}(x - (-1)) \][/tex]
[tex]\[ y + 1 = -\frac{5}{4}(x + 1) \][/tex]
Simplifying further,
[tex]\[ y + 1 = -\frac{5}{4}x - \frac{5}{4} \][/tex]
[tex]\[ y = -\frac{5}{4}x - \frac{5}{4} - 1 \][/tex]
[tex]\[ y = -\frac{5}{4}x - \frac{5}{4} - \frac{4}{4} \][/tex]
[tex]\[ y = -\frac{5}{4}x - \frac{9}{4} \][/tex]
So, the equation of the tangent line is:
[tex]\[ y = -\frac{5}{4}x - \frac{9}{4} \][/tex]
Therefore, the tangent line equation is [tex]\[ \boxed{y = -\frac{5}{4}x - \frac{9}{4}} \][/tex].
Part (a): Find the derivative function [tex]\( f' \)[/tex] for the function [tex]\( f \)[/tex].
The function provided is [tex]\( f(x) = \frac{4}{5x + 1} \)[/tex].
The goal is to find the derivative [tex]\( f'(x) \)[/tex]. The derivative of a function [tex]\( \frac{a}{b(x)} \)[/tex] can be found using the quotient rule or by recognizing it as a constant times the derivative of [tex]\( b(x)^{-1} \)[/tex].
For [tex]\( f(x) = \frac{4}{5x + 1} \)[/tex], we have:
[tex]\[ f(x) = 4 \cdot (5x + 1)^{-1} \][/tex]
Using the chain rule, where [tex]\( g(x) = (5x + 1)^{-1} \)[/tex], we have:
[tex]\[ g(x) = (5x + 1)^{-1} \][/tex]
[tex]\[ g'(x) = -\frac{1}{(5x + 1)^2} \cdot \frac{d}{dx}(5x + 1) \][/tex]
[tex]\[ \frac{d}{dx}(5x + 1) = 5 \][/tex]
So,
[tex]\[ g'(x) = -\frac{5}{(5x + 1)^2} \][/tex]
Thus,
[tex]\[ f'(x) = 4 \cdot g'(x) \][/tex]
[tex]\[ f'(x) = 4 \cdot \left( -\frac{5}{(5x + 1)^2} \right) \][/tex]
[tex]\[ f'(x) = -\frac{20}{(5x + 1)^2} \][/tex]
Therefore,
[tex]\[ f'(x) = -\frac{20}{(5x + 1)^2} \][/tex]
So,
[tex]\[ f'(x) = -\boxed{\frac{20}{(5x + 1)^2}} \][/tex]
Part (b): Determine an equation of the line tangent to the graph of [tex]\( f \)[/tex] at [tex]\( (a, f(a)) \)[/tex] for the given value of [tex]\(a = -1 \)[/tex].
First, we'll find [tex]\( f(a) \)[/tex]. With [tex]\( a = -1 \)[/tex]:
[tex]\[ f(a) = f(-1) = \frac{4}{5(-1) + 1} \][/tex]
[tex]\[ f(-1) = \frac{4}{-5 + 1} \][/tex]
[tex]\[ f(-1) = \frac{4}{-4} \][/tex]
[tex]\[ f(-1) = -1 \][/tex]
We already found the derivative function [tex]\( f'(x) \)[/tex]. Now, we need to compute [tex]\( f'(a) \)[/tex] for [tex]\( a = -1 \)[/tex]:
[tex]\[ f'(-1) = -\frac{20}{(5(-1) + 1)^2} \][/tex]
[tex]\[ f'(-1) = -\frac{20}{(-5 + 1)^2} \][/tex]
[tex]\[ f'(-1) = -\frac{20}{(-4)^2} \][/tex]
[tex]\[ f'(-1) = -\frac{20}{16} \][/tex]
[tex]\[ f'(-1) = -\frac{5}{4} \][/tex]
We now have the slope of the tangent line at [tex]\( x = -1 \)[/tex]:
[tex]\[ m = f'(-1) = -\frac{5}{4} \][/tex]
The equation of the tangent line can be formulated using the point-slope form of the equation of a line:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
With [tex]\( (x_1, y_1) = (-1, -1) \)[/tex]:
[tex]\[ y - (-1) = -\frac{5}{4}(x - (-1)) \][/tex]
[tex]\[ y + 1 = -\frac{5}{4}(x + 1) \][/tex]
Simplifying further,
[tex]\[ y + 1 = -\frac{5}{4}x - \frac{5}{4} \][/tex]
[tex]\[ y = -\frac{5}{4}x - \frac{5}{4} - 1 \][/tex]
[tex]\[ y = -\frac{5}{4}x - \frac{5}{4} - \frac{4}{4} \][/tex]
[tex]\[ y = -\frac{5}{4}x - \frac{9}{4} \][/tex]
So, the equation of the tangent line is:
[tex]\[ y = -\frac{5}{4}x - \frac{9}{4} \][/tex]
Therefore, the tangent line equation is [tex]\[ \boxed{y = -\frac{5}{4}x - \frac{9}{4}} \][/tex].
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
