Finding a Tangent Line of y = arctan(x/2) - (1/(2(x^2 + 4)))

[kinerry]

The problem is laid out like this:

Find the equation of the tangent line to the function at the given point

y = arctan(x/2) - (1 / (2(x^2 + 4))), (0,-1/8)

 

[galactus]

What exactly is it you need to do?. Please give the full problem.

I assume you need to find the tangent line to the given curve at the given coordinates?.

 

[pka]

Re: Finding a Tangent Line

y = arctan(x/2)-(1/(2(x^2+2))), (0,-1/8)

There is something wrong with the problem.
(0,-1/8) is not on the graph: (0,-1/4) is.

 

[kinerry]

Problem corrected

 

[skeeter]

things you need for a tangent line equation ...

1. the point of tangency (x₁, y₁) on the curve ... this was given to you.

2. the slope of the curve at that point of tangency ... you get this by evaluating the value of the function's derivative at that point of tangency, m = y'(x₁).

once you have both, substitute your values into the point-slope form of a linear equation ... y - y₁ = m(x - x₁).

piece of cake, right?

 

[pka]

\(\displaystyle \L
\begin{array}{rcl}
y & = & \arctan \left( {\frac{x}{2}} \right) - \frac{1}{{2\left( {x^2 + 4} \right)}} \\
y' & = & \frac{{\frac{1}{2}}}{{1 + \left( {\frac{x}{2}} \right)^2 }} + \frac{x}{{\left( {x^2 + 4} \right)^2 }} \\
& = & \frac{2}{{4 + x^2 }} + \frac{x}{{\left( {x^2 + 4} \right)^2 }} \\
& = & \frac{{2x^2 + x + 8}}{{\left( {x^2 + 4} \right)^2 }} \\
\end{array}\)


Now the slope is \(\displaystyle y'(0).\)

 

[kinerry]

y - (-1/8) = 0 (x-0)

y = -1/8

 

[pka]

y - (-1/8) = 0 (x-0)

y = -1/8

The slope is NOT zero!
Try again.