If f(x)=x(x-1)(x-4)^2, use interval notation to give all values of x where f(x)>0. Find the critical values by setting all the factors of numerators and denominators = 0 x = 0, x-1 = 0, x-4 = 0 gives critical values 0, 1 and 4 Mark these on a number line 覧覧覧覧熔覧熔覧覧覧覧覧熔覧覧覧覧 -2 -1 0 1 2 3 4 5 6 Choose any value left of 0, say -1 and find f(-1) f(x) = x(x-1)(x-4)ｲ f(-1) = -1(-1-1)(-1-4)ｲ f(-1) = -1(-2)(-5)ｲ f(-1) = +50 We don't care about the value of +50, but only about the fact that it's positive, so we place positive signs above the number line left of 0: + + + + + 覧覧覧覧熔覧熔覧覧覧覧覧熔覧覧覧覧 -2 -1 0 1 2 3 4 5 6 Next we choose any value of x between 0 and 1, say .5 and find f(.5) f(x) = x(x-1)(x-4)ｲ f(.5) = .5(.5-1)(.5-4)ｲ f(-1) = .5(-.5)(-3.5)ｲ f(-1) = -3.0625 We don't care about the value of -3.0625, but only about the fact that it's negative, so we place negative signs above the number line between 0 and 1: + + + + + - - 覧覧覧覧熔覧熔覧覧覧覧覧熔覧覧覧覧 -2 -1 0 1 2 3 4 5 6 Next we choose any value of x between 1 and 4, say 2, and find f(2) f(x) = x(x-1)(x-4)ｲ f(2) = 2(2-1)(2-4)ｲ f(2) = 2(1)(-2)ｲ f(2) = +8 We don't care about the value of +8, but only about the fact that it's positive, so we place positive signs above the number line between 1 and 4: + + + + + - - + + + + + + 覧覧覧覧熔覧熔覧覧覧覧覧熔覧覧覧覧 -2 -1 0 1 2 3 4 5 6 Next we choose any value of x to the right of 4, say 5, and find f(5) f(x) = x(x-1)(x-4)ｲ f(5) = 5(5-1)(5-4)ｲ f(2) = 5(4)(1)ｲ f(2) = +20 We don't care about the value of +20, but only about the fact that it's positive, so we place positive signs above the number line between 1 and 4: + + + + + - - + + + + + + + + + + 覧覧覧覧熔覧熔覧覧覧覧覧熔覧覧覧覧 -2 -1 0 1 2 3 4 5 6 Now we notice that the original problem was to find all values of x for which f(x) > 0, which means that f(x) is positive, so we are to write the interval notation for the part of the number line which has positive signs above it, namely this part: + + + + + - - + + + + + + + + + + <=========o覧熔===========o========> -2 -1 0 1 2 3 4 5 6 The solution will not include the critical points since the inequality is > 0 and not >0. Answer in interval notation: (-oo, 0) U (1,4) U (4,oo) where oo is the infinity symbol. Edwin