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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].
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