Question

A) (x^2-25)(x-6) < 0 b) x^3-49x > 0c) (x^2+8) (x-4) (x-2) < 0d) (x-1) (x-2) (x-3)^2 > 0

Answer 1

Объяснение:

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Answer 2

To solve these inequalities, we'll find the critical points where the expressions equal zero and then test the intervals between these critical points to determine the signs of the expressions. We'll start with each inequality one by one:

a) $(x^2 - 25)(x - 6) < 0$

1. Find the critical points by setting each factor equal to zero: $(x^2 - 25) = 0$ when $x = 5$ or $x = -5$. $(x - 6) = 0$ when $x = 6$.

2. Now, create intervals using these critical points: $(-∞, -5)$, $(-5, 5)$, $(5, 6)$, and $(6, +∞)$.

3. Pick a test point from each interval and evaluate the expression in that interval to determine its sign:

   • For $x < -5$, pick $x = -6$: $((-6)^2 - 25)(-6 - 6) = (36 - 25)(-12) = 11 * (-12) = -132$
   • For $-5 < x < 5$, pick $x = 0$: $(0^2 - 25)(0 - 6) = (-25)(-6) = 150$
   • For $5 < x < 6$, pick $x = 5.5$: $(5.5^2 - 25)(5.5 - 6) = (30.25 - 25)(-0.5) = 5.25 * (-0.5) = -2.625$
   • For $x > 6$, pick $x = 7$: $(7^2 - 25)(7 - 6) = (49 - 25)(1) = 24 * 1 = 24$

4. Now, determine the sign of the expression in each interval:

   • $(-∞, -5)$: Negative
   • $(-5, 5)$: Positive
   • $(5, 6)$: Negative
   • $(6, +∞)$: Positive

5. The inequality is satisfied when the expression is less than zero ($< 0$), which occurs in the intervals where it's negative. Therefore, the solution to $a$ is:

   $(-\infty, -5) \cup (5, 6)$

b) $x^3 - 49x > 0$

1. Find the critical points by setting the expression equal to zero: $x^3 - 49x = 0$ Factor out $x$: $x(x^2 - 49) = 0$ $x(x - 7)(x + 7) = 0$ So, the critical points are $x = -7$, $x = 0$, and $x = 7$.

2. Create intervals using these critical points: $(-∞, -7)$, $(-7, 0)$, $(0, 7)$, and $(7, +∞)$.

3. Pick a test point from each interval and evaluate the expression in that interval to determine its sign:

   • For $x < -7$, pick $x = -8$: $(-8)^3 - 49(-8) = -512 + 392 = -120$
   • For $-7 < x < 0$, pick $x = -1$: $(-1)^3 - 49(-1) = -1 + 49 = 48$
   • For $0 < x < 7$, pick $x = 1$: $(1)^3 - 49(1) = 1 - 49 = -48$
   • For $x > 7$, pick $x = 8$: $(8)^3 - 49(8) = 512 - 392 = 120$

4. Determine the sign of the expression in each interval:

   • $(-∞, -7)$