How do you find the slope of a tangent line to the graph of the function #f(x)=3/x^2 # at (1,3)?

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Answer 1 · 2016-11-23T04:49:19

#y=-6x+9#

The value of the derivative at #x=1# will give us the slope of the tangent line there.

To find the derivative, rewrite the function first.

#f(x)=3x^-2#

Now use the product rule, which states that if #f(x)=ax^n# then #f'(x)=anx^(n-1)#. So

#f'(x)=3(-2)x^(-2-1)=-6x^(-3)=(-6)/x^3#

So the slope of the tangent line at #x=1# is

#f'(1)=(-6)/1^3=-6#

We can relate a line that passes through #(x_1,y_1)# and has slope #m# using the equation

#y-y_1=m(x-x_1)#

With a point of #(1,3)# and slope of #-6# this becomes

#y-3=-6(x-1)#

All on the same side:

#y=-6x+9#

Graphing the function and the tangent line, we can see if we're right:

graph{(y-3/x^2)(-y-6x+9)=0 [-1, 4, -1.044, 7.846]}