Find equation of tangent line to the function f(x)=x^-5/2 at x=3

Expert Answers
lemjay eNotes educator| Certified Educator

First, solve for the corresponding value of  y when x=3.



Apply the negative exponent rule which is a^(-m)=1/a^m.


To simplify further, apply the product rule of exponent which is `a^m*a^n= a^(m+n)` .



Also, express the fractional exponent as radical. And then, rationalize the denominator.

`y= 1/(9*sqrt3)=1/(9*sqrt3)*sqrt3/sqrt3=sqrt3/27`

So the line is tangent to `f(x)=x^(-5/2)` at point `(3,sqrt3/27)` .

Next, solve for the slope of the tangent line. To do so, take the derivative of f(x). And, plug-in x=3.


To simplify, do the same steps as above.

`f'(x)=-5/2*1/3^(7/2)=-5/2*1/(3^(6/2)*3^(1/2)) = -5/2*(1/27*1/(3^(1/2)))`


So, the slope of the tangent line is `-(5sqrt3)/162` .

Now that the point of tangency and slope are known, use the point-slope form to get the equation of tangent line.


Plug-in `m=-(5sqrt3)/162` , `x_1=3` and `y_1=sqrt3/27`


`y-sqrt3/27=-(5sqrt3)/162x +(5sqrt3)/54`

Add both sides by `sqrt3/27` .

`y-sqrt3/27+sqrt3/27=-(5sqrt3)/162x +(5sqrt3)/54+sqrt3/27`

`y=-(5sqrt3)/162x +(5sqrt3)/54+sqrt3/27`

Express `sqrt3/27` with a denominator of 54.  To do so, multiply it by 2/2.

`y=-(5sqrt3)/162x +(5sqrt3)/54+sqrt3/27*2/2`

`y=-(5sqrt3)/162x +(5sqrt3)/54+(2sqrt3)/54`

`y=-(5sqrt3)/162x +(7sqrt3)/54`

Hence the equation of tangent line is  `y=-(5sqrt3)/162x +(7sqrt3)/54` .