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f(x)=(2x-1)(3x-5)(x-4)(3x+6) has zeros at
x=-2,x=(1)/(2),x=(5)/(3), and
x=4.
What is the sign of
f on the interval
(1)/(2) < x < (5)/(3) ?
Choose 1 answer:
(A)
f is always positive on the interval.
(B)
f is always negative on the interval.
(C)
f is sometimes positive and sometimes negative on the interval.

$f(x)=(2 x-1)(3 x-5)(x-4)(3 x+6)$ has zeros at $x=-2, x=\frac{1}{2}, x=\frac{5}{3}$ , and $x=4$ . $\newline$ What is the sign of $f$ on the interval $\frac{1}{2}<x<\frac{5}{3}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $f$ is always positive on the interval. $\newline$ (B) $f$ is always negative on the interval. $\newline$ (C) $f$ is sometimes positive and sometimes negative on the interval.

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Full solution

Q.

f(x)=(2 x-1)(3 x-5)(x-4)(3 x+6)

has zeros at

x=-2, x=\frac{1}{2}, x=\frac{5}{3}

, and

x=4

.

\newline

What is the sign of

f

on the interval

\frac{1}{2}<x<\frac{5}{3}

?

\newline

Choose

1

answer:

\newline

(A)

f

is always positive on the interval.

\newline

(B)

f

is always negative on the interval.

\newline

(C)

f

is sometimes positive and sometimes negative on the interval.

Identify Test Interval: Since the zeros of the function $f(x)$ are at $x=-2$ , $x=\frac{1}{2}$ , $x=\frac{5}{3}$ , and $x=4$ , we can test the sign of $f(x)$ between the zeros by picking a test point in the interval $(\frac{1}{2}, \frac{5}{3})$ .
Choose Test Point: Choose a test point $x=1$ , which is between $\frac{1}{2}$ and $\frac{5}{3}$ .
Determine Sign of Factors: Plug $x=1$ into each factor of $f(x)$ to determine the sign of each factor at $x=1$ .
$(2\cdot1-1) = 1$ , which is positive.
$(3\cdot1-5) = -2$ , which is negative.
$(1-4) = -3$ , which is negative.
$(3\cdot1+6) = 9$ , which is positive.
Multiply Signs: Now, multiply the signs of each factor to find the sign of $f(x)$ at $x=1$ . $\text{Positive} \times \text{Negative} \times \text{Negative} \times \text{Positive} = \text{Positive}.$
Final Conclusion: Since $f(x)$ is positive at $x=1$ , and there are no zeros between $(\frac{1}{2})$ and $(\frac{5}{3})$ , $f(x)$ is always positive on the interval $(\frac{1}{2}, \frac{5}{3})$ .

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