# What is the equation of the line normal to f(x)=4/(2x-1)  at x=0?

Jul 20, 2016

A normal line is a line that is perpendicular to the tangent. In other words, we must first find the equation of the tangent.

#### Explanation:

Step 1: Determine which point the function and the tangent pass through

$f \left(0\right) = \frac{4}{2 \times 0 - 1}$

$f \left(0\right) = \frac{4}{- 1}$

$f \left(0\right) = - 4$

$\therefore$ The function passes through $\left(0 , - 4\right)$

Step 2: Differentiate the function

Let $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$.

Then, $g \left(x\right) = 4$ and $h \left(x\right) = 2 x - 1$. The derivative is given by $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{g ' \left(x\right) \times h \left(x\right) - g \left(x\right) \times h ' \left(x\right)}{h \left(x\right)} ^ 2$

The derivative of $g \left(x\right) = 4$ is $g ' \left(x\right) = 0$. The derivative of $h \left(x\right) = 2 x - 1$ is $2$.

We can now use the quotient rule, as shown above, to determine the derivative.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{g ' \left(x\right) \times h \left(x\right) - g \left(x\right) \times h ' \left(x\right)}{h \left(x\right)} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{0 \times \left(2 x - 1\right) - 4 \left(2\right)}{2 x - 1} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{8}{2 x - 1} ^ 2$

Step 3: Determine the slope of the tangent

The slope of the tangent is given by evaluating $f \left(a\right)$ inside the derivative, where $x = a$ is the given point.

Then, we can say:

${m}_{\text{tangent}} = - \frac{8}{2 \times 0 - 1} ^ 2 = \frac{- 8}{1} = - 8$

Step 4: Determine the slope of the normal line

As mentioned earlier, the normal line is perpendicular, but passes through the same point of tangency that does the tangent. A line perpendicular to another has a slope that is the negative reciprocal of the other. The negative reciprocal of $- 8$ is $\frac{1}{8}$. Thus, the slope of the normal line is $\frac{1}{8}$.

Step 5: Determine the equation of the normal line using point-slope form

We now know the slope of the normal line as well as the point of contact. This is enough for us to determine its equation using point-slope form.

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - \left(- 4\right) = \frac{1}{8} \left(x - 0\right)$

$y + 4 = \frac{1}{8} x - 0$

$y = \frac{1}{8} x - 4$

In summary...

The line normal to $f \left(x\right) = \frac{4}{2 x - 1}$ at the point $x = 0$ has an equation of $y = \frac{1}{8} x - 4$.

Practice exercises:

Determine the equation of the normal lines to the given relations at the indicated point.

a) $f \left(x\right) = \frac{2 {x}^{2} + 6}{3 {x}^{2} - 3}$, at $x = 3$

b) $f \left(x\right) = {e}^{{x}^{2} - 4} + 2 {x}^{7}$, at the point $x = - 2$

c) $f \left(\theta\right) = \sin 2 \theta$, at the point $\theta = \frac{\pi}{4}$

d) $f \left(x\right) = {\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2} = 53$ at the point $\left(9 , - 1\right)$

Hopefully this helps, and good luck!