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

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### 1 Answer

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

`y=f(x)`

`y=x^(-5/2)=3^(-5/2)`

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

`y=1/3^(5/2)`

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

`y=1/(3^(4/2)*3^(1/2))`

`y=1/(9*3^(1/2))`

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.

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

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)))`

`f'(x)=-5/(54sqrt3)=-5/(54sqrt3)*sqrt3/sqrt3=-(5sqrt3)/162`

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.

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

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

`y-sqrt3/27=-(5sqrt3)/162(x-3)`

`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` .**