Question

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

Answer 1

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(0) = 4/(2 xx 0 - 1)`

`f(0) = 4/(-1)`

`f(0) = -4`

`:.` The function passes through `(0, -4)`

Step 2: Differentiate the function

Let `f(x) = (g(x))/(h(x))`.

Then, `g(x) = 4` and `h(x) = 2x - 1`. The derivative is given by `dy/dx = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2`

The derivative of `g(x) = 4` is `g'(x) = 0`. The derivative of `h(x) = 2x - 1` is `2`.

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

`dy/dx = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2`

`dy/dx = (0 xx (2x - 1) - 4(2))/(2x - 1)^2`

`dy/dx = -8/(2x - 1)^2`

Step 3: Determine the slope of the tangent

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

Then, we can say:

`m_"tangent" = -8/(2 xx 0 - 1)^2 = (-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 `1/8`. Thus, the slope of the normal line is `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(x - x_1)`

`y - (-4) = 1/8(x - 0)`

`y + 4 = 1/8x - 0`

`y = 1/8x - 4`

In summary...

The line normal to `f(x) = 4/(2x - 1)` at the point `x = 0` has an equation of `y = 1/8x - 4`.

Practice exercises:

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

a) `f(x) = (2x^2 + 6)/(3x^2 - 3)`, at `x = 3`

b) `f(x) = e^(x^2 - 4) + 2x^7`, at the point `x = -2`

c) `f(theta) = sin2theta`, at the point `theta = pi/4`

d) `f(x) = (x - 2)^2 + (y + 3)^2 =53` at the point `(9, -1)`

Hopefully this helps, and good luck!