I'd like help with all of the first exercise, and then the two last parts of the last two exercises, 'cause I've done the first parts.
Hope that made sense.
Thanks in advance (:
Find the equation of the normal line to
$\displaystyle y = \frac{1}{(x^2 + 1)^2}$
at the point
$\displaystyle \left ( 1, \frac{1}{4} \right )$
First we want the value of the slope of the tangent line at this point. So:
$\displaystyle \frac{dy}{dx} = -\frac{4x}{(x^2 + 1)^3}$
At x = 1 this is
$\displaystyle \frac{dy}{dx}_{x = 1} = -\frac{1}{2}$
Now, the line normal to the curve at this point has a slope of $\displaystyle -\frac{1}{-\frac{1}{2}} = 2$
So it has the form
$\displaystyle y = 2x + b$
We know this line passes through the point $\displaystyle
\left ( 1, \frac{1}{4} \right )$, so
$\displaystyle \frac{1}{4} = 2 \cdot 1 + b$
$\displaystyle b = -\frac{7}{4}$
Thus the normal line is $\displaystyle y = 2x - \frac{7}{4}$
-Dan
Come now! The second part of number 3 is not a Calculus problem.
For what values of x is
$\displaystyle \frac{dy}{dx} = \frac{x^2 + 4x - 7}{(x + 2)^2}$
0 or undefined.
For the first part solve
$\displaystyle \frac{x^2 + 4x - 7}{(x + 2)^2} = 0$
For the second part, consider what values of x make the denominator 0 (which is forbidden.)
-Dan