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

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Answer 1 · 2018-07-14T17:09:15

#y=-1/2x+11/10#

The equation of the Tangent line has the form #y=mx+n#

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

#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#