f(x) = (x - 5)(x - 1)/(x - 3)(x - 2)
Now note that the graph of the denominator is steeper than the graph of the numerator.
So, now put the intervals given in the question. You'll get the results. [1]
f(x) = (x - 5)(x - 1)/(x - 3)(x - 2)
Now note that the graph of the denominator is steeper than the graph of the numerator.
So, now put the intervals given in the question. You'll get the results. [1]
Use the wavy-curve method, its pretty easy..
Mark the points where f(x) becomes zero.
for example,
f(x) = (x - 5)(x - 1) = x2 - 6x + 5
Check the sign of the coefficient of x2, if it is negative, then the graph approaches negative infinity as x reaches infinity and vice versa.
Keep in mind the value of f(x) in different intervals and plot the graph accordingly.
as a shortcut , u can also check by putting values
:P
with no offence to the purists
by putting values, if suppose f(x)=-3.
Then how can u verify if it satisfies both f(x) less than 0 and-1.
So plz giv me a subjective ans