Answer to Question #245304 in Algebra for minu

Question

Let f(x)=x^3-6x^2+3x+10f(x)=x3−6x2+3x+10, then choose the set of correct options regarding f(x).

1. If x \in [0, 1] \cup (10, \infty)x∈[0,1]∪(10,∞), then f(x)is positive.
2. If x \in (-\infty, -1] \cup (3, 6)x∈(−∞,−1]∪(3,6), then f(x) is negative.
3. If x \in (-\infty, 1] \cup (2, 5)x∈(−∞,1]∪(2,5), then f(x) is negative.
4. If x \in [-2, 2] \cup (5, \infty)x∈[−2,2]∪(5,∞), then f(x) is positive.
5. If x \in (-1, 2) \cup (5, \infty)x∈(−1,2)∪(5,∞), then f(x)is positive.

Answer 1

$$f(x)=0=>x^3-6x^2+3x+10=0$$

$$x^2(x+1)-7x(x+1)+10(x+1)=0$$

$$(x+1)(x^2-7x+10)=0$$

$$(x+1)(x-2)(x-5)=0$$

$$x_1=-1, x_2=2, x_3=5$$

If $x<-1,$ then $f(x)<0.$

If $-1<x<2,$ then $f(x)>0.$

If $2<x<5,$ then $f(x)<0.$

If $x>5,$ then $f(x)>0.$

Answer:

5. If $x\in(-1, 2)\cup(5,\infin),$ then $f(x)$ is positive.