Question

what is the range of the function v(x) = x(9 - x)(14 - 2x)?
a 0 ≤ y ≤ 146.432
b 0 < y < ∞
c -∞ < y < ∞
d 0 < y ≤ 146.432

Explanation:

Step1: Analyze the function's context

The function \( V(x)=x(9 - x)(14 - 2x) \) likely represents a volume (e.g., of a box with side - related dimensions), so \( x>0 \), \( 9 - x>0\Rightarrow x < 9 \), and \( 14-2x>0\Rightarrow x < 7 \). So the domain of \( x \) is \( 0 < x < 7 \) (since \( x>0 \) and the most restrictive upper bound is \( x < 7 \) from \( 14 - 2x>0 \)). But when \( x = 0 \) or \( x=7 \) or \( x = 9 \), \( V(x)=0 \).

Step2: Analyze the type of function

\( V(x)=x(9 - x)(14 - 2x)=x(126-18x - 14x + 2x^{2})=2x^{3}-32x^{2}+126x \). This is a cubic function. For a cubic function \( ax^{3}+bx^{2}+cx + d\) (\( a = 2
eq0 \)), as \( x
ightarrow\pm\infty \), the function will tend to \( \pm\infty \), but in the context of volume (if it is a volume function), the domain of \( x \) is restricted to non - negative values that make the dimensions positive. However, when we expand the function \( V(x)=x(9 - x)(14 - 2x) \), we can also analyze its critical points. First, find the derivative \( V^\prime(x)=(9 - x)(14 - 2x)+x(- 1)(14 - 2x)+x(9 - x)(-2)\). Simplify \( V^\prime(x)=(14 - 2x)(9 - x - x)+x(-2)(9 - x)= (14 - 2x)(9 - 2x)-2x(9 - x)\). Expand: \( V^\prime(x)=126-28x-18x + 4x^{2}-18x + 2x^{2}=6x^{2}-64x + 126 \). Set \( V^\prime(x)=0 \), \( 3x^{2}-32x + 63 = 0 \). Using the quadratic formula \( x=\frac{32\pm\sqrt{32^{2}-4\times3\times63}}{2\times3}=\frac{32\pm\sqrt{1024 - 756}}{6}=\frac{32\pm\sqrt{268}}{6}=\frac{32\pm2\sqrt{67}}{6}=\frac{16\pm\sqrt{67}}{3}\approx\frac{16\pm8.19}{3}\). So \( x_1\approx\frac{16 + 8.19}{3}\approx8.06\) (but this is outside the domain \( 0

Answer:

A. \( 0\leq y\leq146.432 \)