How do you find the domain and range of #f(x)=(x^2+5x-6)/(x^2-5x+6)#?

Question

Nathan Axel · Accepted Answer

The domain is #RR-{2,3}#
The range is #(-oo,-49]uu[-1,+oo)# For the domain, the denominator must be #!=0# Consequently, The domain is #RR-{2,3}# Let us rewrite the function for the range. #y=(x^2+5x-6)/(x^2-5x+6)# #y(x^2-5x+6)=x^2+5x-6# #yx^2-x^2-5yx-5x+6y+6=0# #(y-1)x^2-5(y+1)x+6(y+1)=0#....................#(1)# Let 's calculate the discriminant of equation #(1)# #Delta>=0# #25(y+1)^2-24(y-1)(y+1)>=0# #25(y^2+2y+1)-24(y^2-1)>=0# #y^2+50y+49>=0# #(y+49)(y+1)>=0# The range is #(-oo,-49]uu[-1,+oo)# graph{[-132.6, 134.3, -94.2, 39.3]} = (x^2+5x-6)/(x^2-5x+6)

Nathan Axel · Answer

undefined To find the domain of $ f(x) = \frac{x^2 + 5x - 6}{x^2 - 5x + 6} $, we need to determine the values of $ x $ for which the function is defined. The function is undefined when the denominator is equal to zero because division by zero is undefined.

Therefore, we need to find the values of $ x $ that make $ x^2 - 5x + 6 = 0 $. Factoring the denominator gives us $ (x - 2)(x - 3) = 0 $, so $ x = 2 $ or $ x = 3 $. These values make the denominator zero, so they are not in the domain.

Therefore, the domain of $ f(x) $ is all real numbers except $ x = 2 $ and $ x = 3 $.

To find the range, we need to determine the possible values of $ f(x) $ for the values of $ x $ in the domain. Since $ f(x) $ is a rational function, its range will be all real numbers except for the values that make the denominator zero. Therefore, the range of $ f(x) $ is all real numbers except for the values that make $ x^2 - 5x + 6 = 0 $, which are $ x = 2 $ and $ x = 3 $.