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secondary 3 | A Maths
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Pls help me continue from here
Differentiating 1 gets us 0.
Anyway, you will get dy/dx = -1 divided by (x - 1)^2.
Note that the curve is not defined for x = 1 since y = 1/(x - 1) = 1/0 has no value.
For all other values of x, (x - 1)^2 will always be a positive value. This is an important property - the output of a squared function is always non-negative.
But the numerator in dy/dx contains a negative sign, so the entire expression -1 divided by (x - 1)^2 is going to be negative in value (negative/positive = negative).
As such, for x < 1, dy/dx is negative…
and for x > 1, dy/dx is also negative.
A negative value for dy/dx indicates that the curve is decreasing in a given interval.
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