## To find the equation of the normal to the graph of f(x) = x / (2x-3) at the point (1, f(1)), we need to first find the derivative of f(x), which is denoted as f'(x).

Differentiating f(x) with respect to x using the quotient rule, we get:

f'(x) = [(2x-3)(1) - x(2)] / (2x-3)^2

Simplifying this expression, we have:

f'(x) = (2x - 3 - 2x) / (2x - 3)^2

f'(x) = (-3) / (2x - 3)^2

Now, to find the slope of the normal to the graph at the point (1, f(1)), we can substitute x = 1 into f'(x):

f'(1) = (-3) / (2(1) - 3)^2

f'(1) = (-3) / (-1)^2

f'(1) = -3

Therefore, the slope of the normal to the graph at (1, f(1)) is -1/f'(1), which is -1/-3 = 1/3.

Now, we can determine the equation of the normal using point-slope form. The equation will be of the form g(x) = mx + b, where m is the slope of the normal and b is the y-intercept of the normal.

Substituting the slope, m = 1/3, into the equation, we have:

g(x) = (1/3)x + b

To find b, we can substitute the coordinates of the given point (1, f(1)) into the equation and solve for b.

So, we have:

f(1) = (1/3)(1) + b

Simplifying further, we get:

f(1) = 1/3 + b

Now, solve for b:

f(1) - 1/3 = b

Finally, substitute the value of b back into the equation to get the equation of the normal:

g(x) = (1/3)x + (f(1) - 1/3)

Therefore, the equation of the normal to the graph of f(x) = x / (2x-3) at the point (1, f(1)) is g(x) = (1/3)x + (f(1) - 1/3).