Question

1)y=4x^3/2 2)f(x)=(3x^2+1)(2x^2+3)

Answer 1

1)
y=4x³:2
y=64x:2
y=32x

2)
f(x)=(3x²+1)×(2x²+3)
f(x)=(9x+1)×(4x+3)

Answer 2

To analyze the given functions, let's break down each one and discuss their properties.

1. $y = \frac{4x^3}{2}$

This function can be simplified to:

$y = 2x^3$

Properties:

• Polynomial Function: It is a polynomial function of degree 3, which means the highest power of $x$ in the equation is 3.
• Symmetry: The function is not symmetric about the y-axis or the origin since all powers of $x$ are odd (there are no even powers).
• Increasing/Decreasing: As the leading coefficient (2) is positive, the function is increasing for positive values of $x$ and decreasing for negative values of $x$.
• Inflection Points: The function has one inflection point, as odd-degree polynomials have at most one inflection point.
• Roots: To find the roots (where $y = 0$), set $2x^3 = 0$, which gives $x = 0$. So, the function intersects the x-axis at the point (0, 0).

2. $f(x) = (3x^2 + 1)(2x^2 + 3)$

Properties:

• Polynomial Function: It is a polynomial function of degree 4, as it is the product of two polynomials, each of degree 2.
• Symmetry: The function is not symmetric about the y-axis or the origin.
• Increasing/Decreasing: To determine the intervals of increasing and decreasing, we need to analyze the sign of $f'(x)$, the derivative of $f(x)$.
  • First, find $f'(x)$: $f'(x) = 2(3x^2 + 3) + 3(2x^2 + 1) = 12x^2 + 12x + 3$
  • Set $f'(x) = 0$ to find critical points: $12x^2 + 12x + 3 = 0$ Divide the equation by 3: $4x^2 + 4x + 1 = 0$ This quadratic equation has a double root: $x = -\frac{1}{2}$
  • Now, use the first derivative test to determine increasing/decreasing intervals:
    • For $x < -\frac{1}{2}$, $f'(x) > 0$, so $f(x)$ is increasing.
    • For $x > -\frac{1}{2}$, $f'(x) > 0$, so $f(x)$ is decreasing.
• Inflection Points: To find the inflection points, we need to analyze the second derivative $f''(x)$ for possible points of inflection.
  • First, find $f''(x)$: $f''(x) = 24x + 12$
  • Set $f''(x) = 0$ to find possible inflection points: $24x + 12 = 0$ $x = -\frac{1}{2}$
• Roots: To find the roots of $f(x)$, set $f(x) = 0$: $(3x^2 + 1)(2x^2 + 3) = 0$ This equation has no real roots since both factors are sums of squares, and squares are always non-negative.

Overall, the two functions have different properties. The first function is a simple cubic polynomial with one real root at the origin (0, 0). The second function is a quartic polynomial with no real roots but has one inflection point at $x = -\frac{1}{2}$. It also exhibits both increasing and decreasing behavior on different intervals.