How do you find the equation of the line tangent to the graph of $f (x) = \frac{10}{5 x + 9}$ $f(x) =10/(5x + 9$ ), when x =1/5?

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Answer 1 · 2018-07-14T17:09:15

$y = - \frac{1}{2} x + \frac{11}{10}$ $y=-1/2x+11/10$

The equation of the Tangent line has the form $y = m x + n$ $y=mx+n$

At first we will compute the first derivative of $f (x)$ $f(x)$

$f (x) = 10 {(5 x + 9)}^{- 1}$ $f(x)=10(5x+9)^(-1)$
then we get

$f'(x)=-10(5x+9)^(-2)*5=-50/(5x+9)^2$ by the chain rule.
and

$f'(1/5)=-50/(5*1/5+9)^2=-50/100=-1/2$
so the slope is given by $m=-1/2$

The Point is given by $f(1/5)=10/10=1$

so we have
$1=-1/2*(1/5)+n$

so $n=11/10$

How do you find the equation of the line tangent to the graph of f(x) =10/(5x + 9f(x)=105x+9f(x) =10/(5x + 9), when x =1/5?