Suppose that $f (x) = x^{3} - 3 x^{2} - 4 x + 12$ and $h (x) = [\begin{matrix} \frac{f (x)}{x - 3}, x \neq 3 K, x = 3 \end{matrix}$ ,then(a)find the zeros of $f$
(b)find the value of $k$ that makes $h$ continuous at $x = 3$
(c)Using the value of $k$
found in (b), determine whether $h$ is an even function.

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Solution

$\begin{matrix} h (x) = [\begin{matrix} \frac{f (x)}{x - 3}, x \neq 3 K, x = 3 \end{matrix} = [\begin{matrix} \frac{(x - 2) (x - 3) (x + 2)}{x - 3}, x \neq 3 K, x = 3 \end{matrix} = [\begin{matrix} (x - 2) (x + 2), x \neq 3 K, x = 3 \end{matrix} k = {lim}_{x \to 3^{+}} (x - 2) (x + 2) = (3 - 2) (3 + 2) = 5 \end{matrix}$

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Similar questions

x^{3} - 3 x^{2} - 4 x + 12

⎧ ⎨ ⎩ \begin{matrix} \frac{f (x)}{x - 3} x \neq 3 K x = 3 \end{matrix}

x = 3

f (x) = \frac{x^{2} - 9}{x - 3}, x \neq 3

f (x) = k, x = 3

^{'} k^{'}

x = 3

(x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{{(x + 3)}^{2} - 36}{x - 3}, & x \neq 3 k, & x = 3 \end{matrix}

x = 3

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{(x + 3)^{2} - 36}{x - 3} & , x \neq 3 k & , x = 3 \end{matrix}

x = 3,

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{(x + 3)^{2} - 36}{x - 3}, & x \neq 3 k, & x = 3 \end{matrix}

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