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Determine the value of [tex]f(-3)[/tex] for the function

[tex]\[ f(x) = \begin{cases}
x^3 + 18x^2 + 104x + 192 & \text{for} & x \ \textless \ -3 \\
\frac{3x + 9}{x^3 - 4x^2 + x - 6} & \text{for} & -3 \leq x \ \textless \ 3 \\
\sqrt{x^2 - 9} & \text{for} & x \geq 3
\end{cases} \][/tex]

A. 0
B. [tex]\(\frac{3}{8}\)[/tex]
C. Undefined
D. 15

Sagot :

GinnyAnsweridn
To determine the value of [tex]\( f(-3) \)[/tex] for the given piecewise function:
[tex]\[ f(x) = \begin{cases} x^3 + 18x^2 + 104x + 192 & \text{for } x < -3 \\ \frac{3x + 9}{x^3 - 4x^2 + x - 6} & \text{for } -3 \leq x < 3 \\ \sqrt{x^2 - 9} & \text{for } x \geq 3 \end{cases} \][/tex]

Since [tex]\(-3 \leq -3 < 3\)[/tex], we use the second piece of the piecewise function:
[tex]\[ f(x) = \frac{3x + 9}{x^3 - 4x^2 + x - 6} \][/tex]

Substitute [tex]\( x = -3 \)[/tex] into the function:
[tex]\[ f(-3) = \frac{3(-3) + 9}{(-3)^3 - 4(-3)^2 + (-3) - 6} \][/tex]

Simplify the expression in the numerator:
[tex]\[ 3(-3) + 9 = -9 + 9 = 0 \][/tex]

Next, simplify the expression in the denominator:
[tex]\[ (-3)^3 - 4(-3)^2 + (-3) - 6 = -27 - 36 - 3 - 6 = -72 \][/tex]

So, the function simplifies to:
[tex]\[ f(-3) = \frac{0}{-72} = 0 \][/tex]

Therefore, the value of [tex]\( f(-3) \)[/tex] is:
[tex]\[ \boxed{-0.0} \][/tex]
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