# Equation for tangent line for $f(x) = 1/\sqrt{x}$ at $x=a$

first time math stack-exchange-er here.

I'm self-teaching single variable calculus using MIT's free online courses and I think I found a typo in the homework solution set (problem 1C-4 part d). I'm not confident enough in my own abilities to know for sure if this is a mistake vs. my poor math skills.

Could anyone tell me if the following equation is correct?

The problem is as follows: Write an equation for the tangent line for the following functions:

$$f(x) = \frac1{\sqrt{x}}\ \text{ at } x=a$$

I did the following. First I found the derivative of $f(x)$: $$f'(x) = -\frac12 x^{-\frac32}\$$

Then I plugged in a to get $f(a)$ and $f'(a)$ and used the point-slope method to find the equation for the tangent line: $$y-a^{-\frac12} = -\frac12 a^{-\frac32}(x-a)\$$

Which I then simplified to: $$y=-\frac12a^{-\frac32}x + \frac32 a^{-\frac12}\$$

However, the solution set says the answer is: $$y=-a^{-\frac32}x + \frac32 a^{-\frac12}\$$

Is the solution set correct? If so, where did the $-\frac12$ go?

Thank you!

• Welcome to Math.SE! A very excellently asked first question. – Sean Roberson Jul 6 '17 at 21:18
• I get the same answer as you did, seems solution set answer is wrong. – coffeemath Jul 6 '17 at 21:19
• You are right ... there is a mistake in their solutions. – Donald Splutterwit Jul 6 '17 at 21:21
• Don’t forget: the answers to problems are often supplied by underpaid and overworked graduate students. And it would be unheard of for them to be checked by the author(s) of the text. – Lubin Jul 7 '17 at 0:09
• Also, notice another typo in it: the minus sign in $a^{3/2}$ is omitted. – farruhota Jul 7 '17 at 14:02

## 1 Answer

Just to reiterate what was said in the comments and to remove this question from the unanswered queue, your solution is correct and the one provided by the book is wrong.