Precalculus Textbook 8th Edition Section 2.8 Exercise 25 Solution
Question:
Determine the real values of x that causes the function \[ f(x) = \frac{(x-1)}{(2x+3)(x-4)} \]
to be
(a) Zero,
(b) Undefined,
(c) Positive,
(d) Negative
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Solution:
The function \[ f(x) = \frac{(x-1)}{(2x+3)(x-4)} \]
(a) The given function is to be zero. It means, we need to solve for x in equation f(x) = 0. \[ \Rightarrow \frac{(x-1)}{(2x+3)(x-4)}=0 \Rightarrow (x-1)=0 \Rightarrow x=1 \]
(b) The given function is to be undefined where denominator is zero. It means, we need to solve for x in equation (2x + 3)(x – 4) = 0. \[ \Rightarrow (2x+3)(x-4)=0 \Rightarrow x=\frac{-3}{2} and -4 \]
(c) The given function is to be positive. It means, \[ f(x) = \frac{(x-1)}{(2x+3)(x-4)} \ge 0\Rightarrow x\in (-\frac{3}{2},1]\cup (4,\infty)\]
(d) The given function is to be negative. It means, \[ f(x) = \frac{(x-1)}{(2x+3)(x-4)} \le 0\Rightarrow x\in (-\infty,-\frac{3}{2}) \cup [1,4)\]
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