College

Find the transition points for the function [tex]y = 3x^5 - 45x^3[/tex].

1. The critical point(s) at [tex]x = \square[/tex]

2. The point(s) of inflection at [tex]x = \square[/tex]

(Use symbolic notation and fractions where needed. Provide your answers in the form of comma-separated values.)

Answer :

Sure! Let's find the critical points and points of inflection for the function [tex]\( y = 3x^5 - 45x^3 \)[/tex].

### Critical Points

1. Find the First Derivative: The first step is to find the derivative of the function with respect to [tex]\( x \)[/tex]:
[tex]\[
y' = \frac{d}{dx}(3x^5 - 45x^3) = 15x^4 - 135x^2
\][/tex]

2. Set the First Derivative to Zero: Solve the equation [tex]\( y' = 0 \)[/tex] to find the critical points:
[tex]\[
15x^4 - 135x^2 = 0
\][/tex]

3. Factor the Equation: Let's factor the equation:
[tex]\[
15x^2(x^2 - 9) = 0
\][/tex]
This gives us two solutions:
[tex]\[
x^2 = 0 \quad \text{or} \quad x^2 - 9 = 0
\][/tex]
Solving these:
[tex]\[
x = 0 \quad \text{or} \quad x = \pm 3
\][/tex]

4. Critical Points: Therefore, the critical points are at:
[tex]\[
x = -3, 0, 3
\][/tex]

### Points of Inflection

1. Find the Second Derivative: Next, take the derivative of [tex]\( y' \)[/tex] to find the second derivative:
[tex]\[
y'' = \frac{d}{dx}(15x^4 - 135x^2) = 60x^3 - 270x
\][/tex]

2. Set the Second Derivative to Zero: Solve the equation [tex]\( y'' = 0 \)[/tex] to find the points of inflection:
[tex]\[
60x^3 - 270x = 0
\][/tex]

3. Factor the Equation: Factor the second derivative:
[tex]\[
60x(x^2 - \frac{27}{6}) = 0
\][/tex]
Solving this:
[tex]\[
x = 0 \quad \text{or} \quad x = \pm \frac{3\sqrt{2}}{2}
\][/tex]

4. Points of Inflection: The points of inflection are at:
[tex]\[
x = 0, -\frac{3\sqrt{2}}{2}, \frac{3\sqrt{2}}{2}
\][/tex]

### Final Answer
- The critical points are at [tex]\( x = -3, 0, 3 \)[/tex].
- The points of inflection are at [tex]\( x = 0, -\frac{3\sqrt{2}}{2}, \frac{3\sqrt{2}}{2} \)[/tex].

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