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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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` .

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial Team