f(x)=(x-3)(x+2)(x+4)(x-1)(2x-9) has zeros at x=-4,x=-2,x=1,x=3, and x=(9)/(2). What is the sign of f on the interval -2 < x < (9)/(2) ? 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.

Identify Sign Changes: Since the function f(x) is a product of linear factors, the sign of f(x) changes at each zero.Odd Number of Zeros: Between the zeros x = -2 and x = 1, there is an odd number of zeros (1 zero at x = 1), so the sign of f(x) changes once.Sign Change at Zeros: Between the zeros x = 1 and x = 3, there is an odd number of zeros (1 zero at x = 3), so the sign of f(x) changes again.No Zeros Between: Between the zeros x = 3 and x = 9/2, there are no zeros, so the sign of f(x) does not change.Test Value Selection: Since f(x) changes sign at x = 1 and x = 3, we need to test a value between -2 and 9/2 to determine the sign on the interval.Plug in Test Value: Let's pick x = 0, which is between -2 and 9/2, and plug it into f(x) to determine the sign.Determine Sign: f(0) = (0-3)(0+2)(0+4)(0-1)(2*0-9) = (-3)(2)(4)(-1)(-9)Final Sign Analysis: f(0) = -3 * 2 * 4 * -1 * -9 = -216, which is negative.Since f(0) is negative and the sign changes at x = 1 and x = 3, f(x) is negative between -2 and 1, then positive between 1 and 3, and negative again between 3 and 9/2.Therefore, f(x) is sometimes positive and sometimes negative on the interval -2 < x < 9/2.

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f(x)=(x-3)(x+2)(x+4)(x-1)(2x-9) has zeros at
x=-4,x=-2,x=1,x=3, and
x=(9)/(2).
What is the sign of
f on the interval
-2 < x < (9)/(2) ?
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.

$f(x)=(x-3)(x+2)(x+4)(x-1)(2 x-9)$ has zeros at $x=-4, x=-2, x=1, x=3$ , and $x=\frac{9}{2}$ . $\newline$ What is the sign of $f$ on the interval \( -2

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Math Problems

Algebra 1

Power rule with rational exponents

Full solution

f(x)=(x-3)(x+2)(x+4)(x-1)(2 x-9)

has zeros at

x=-4, x=-2, x=1, x=3

, and

x=\frac{9}{2}

\newline

What is the sign of

f

on the interval

-2<x<\frac{9}{2}

\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 Sign Changes: Since the function $f(x)$ is a product of linear factors, the sign of $f(x)$ changes at each zero.
Odd Number of Zeros: Between the zeros $x = -2$ and $x = 1$ , there is an odd number of zeros ( $1$ zero at $x = 1$ ), so the sign of $f(x)$ changes once.
Sign Change at Zeros: Between the zeros $x = 1$ and $x = 3$ , there is an odd number of zeros ( $1$ zero at $x = 3$ ), so the sign of $f(x)$ changes again.
No Zeros Between: Between the zeros $x = 3$ and $x = \frac{9}{2}$ , there are no zeros, so the sign of $f(x)$ does not change.
Test Value Selection: Since $f(x)$ changes sign at $x = 1$ and $x = 3$ , we need to test a value between $-2$ and rac{9}{2} to determine the sign on the interval.
Plug in Test Value: Let's pick $x = 0$ , which is between $-2$ and $\frac{9}{2}$ , and plug it into $f(x)$ to determine the sign.
Determine Sign: $f(0) = (0-3)(0+2)(0+4)(0-1)(2\cdot 0-9) = (-3)(2)(4)(-1)(-9)$
Final Sign Analysis: $f(0) = -3 \times 2 \times 4 \times -1 \times -9 = -216$ , which is negative.
Final Sign Analysis: $f(0) = -3 \times 2 \times 4 \times -1 \times -9 = -216$ , which is negative.Since $f(0)$ is negative and the sign changes at $x = 1$ and $x = 3$ , $f(x)$ is negative between $-2$ and $1$ , then positive between $1$ and $3$ , and negative again between $3$ and $f(0)$ $0$ .
Final Sign Analysis: $f(0) = -3 \times 2 \times 4 \times -1 \times -9 = -216$ , which is negative.Since $f(0)$ is negative and the sign changes at $x = 1$ and $x = 3$ , $f(x)$ is negative between $-2$ and $1$ , then positive between $1$ and $3$ , and negative again between $3$ and $f(0)$ $0$ .Therefore, $f(x)$ is sometimes positive and sometimes negative on the interval $f(0)$ $2$ .