Question

segment 25a - just a few higher - order inequalities

1. lets look at (x - 4)(x - 7)^4>0

a. what are the two critical points here?
b. when testing critical points, a student wishes to use x = 3 to test the left - most region, but really does not wish to calculate the value of (3 - 7)^4. is this calculation required, or is there a faster way to perform this test?
c. find the solution in interval notation.

Explanation:

Step1: Find critical points

Set each factor equal to zero. For $(x - 4)(x - 7)^8>0$, $x-4 = 0$ gives $x = 4$ and $(x - 7)^8=0$ gives $x = 7$.

Step2: Analyze sign - testing

When $x = 3$, we consider the signs of the factors. The factor $x - 4$ is $3-4=-1<0$. The factor $(x - 7)^8$ is always non - negative for all real $x$ (since it's an even - power). When $x<4$, $x - 4<0$ and $(x - 7)^8>0$ for $x
eq7$. We don't need to calculate $(3 - 7)^8$ exactly. We just need to know its sign. Since the exponent 8 is even, $(x - 7)^8>0$ for $x
eq7$.

Step3: Determine solution interval

We want $(x - 4)(x - 7)^8>0$. Since $(x - 7)^8>0$ for $x
eq7$, we need $x - 4>0$. The solution is $x>4$ and $x
eq7$. In interval notation, it is $(4,7)\cup(7,\infty)$.

Answer:

a. The critical points are $x = 4$ and $x = 7$.
b. No, the exact calculation of $(3 - 7)^8$ is not required. We just need to know the sign of the factors. Since the exponent of $(x - 7)$ is even, $(x - 7)^8>0$ for $x
eq7$.
c. $(4,7)\cup(7,\infty)$