f(x)=(x-2)(x+4)(x+5)(x-1)(x-9) has zeros at x=-5,x=-4,x=1,x=2, and x=9. What is the sign of f on the interval -4 < x < 1 ? 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.

Sign Change at Zeros: Since f(x) is a product of linear factors, the sign of f(x) changes at each zero of the function.Consistent Sign Interval: Between the zeros x=-4 and x=1, there are no other zeros, so the sign of f(x) will be consistent in this interval.Determining Sign at Test Point: To determine the sign of f(x) in the interval -4 < x < 1, we can pick a test point in the interval, like x=0.Calculate f(0): Plug x=0 into f(x) to get f(0)=(0-2)(0+4)(0+5)(0-1)(0-9).Final Calculation: Calculate f(0) = (-2)(4)(5)(-1)(-9).Conclusion: f(0) = -2 * 4 * 5 * -1 * -9 = -360, which is negative.Since f(0) is negative and there are no zeros between -4 and 1, f(x) is always negative on the interval -4 < x < 1.

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f(x)=(x-2)(x+4)(x+5)(x-1)(x-9) has zeros at
x=-5,x=-4,x=1,x=2, and
x=9.
What is the sign of
f on the interval
-4 < x < 1 ?
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.

$f(x)=(x-2)(x+4)(x+5)(x-1)(x-9)$ has zeros at $x=-5, x=-4, x=1, x=2$ , and $x=9$ . $\newline$ What is the sign of $f$ on the interval \( -4

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Full solution

f(x)=(x-2)(x+4)(x+5)(x-1)(x-9)

has zeros at

x=-5, x=-4, x=1, x=2

, and

x=9

\newline

What is the sign of

f

on the interval

-4<x<1

\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.

Sign Change at Zeros: Since $f(x)$ is a product of linear factors, the sign of $f(x)$ changes at each zero of the function.
Consistent Sign Interval: Between the zeros $x=-4$ and $x=1$ , there are no other zeros, so the sign of $f(x)$ will be consistent in this interval.
Determining Sign at Test Point: To determine the sign of $f(x)$ in the interval -4 < x < 1, we can pick a test point in the interval, like $x=0$ .
Calculate $f(0)$ : Plug $x=0$ into $f(x)$ to get $f(0)=(0-2)(0+4)(0+5)(0-1)(0-9)$ .
Final Calculation: Calculate $f(0) = (-2)(4)(5)(-1)(-9)$ .
Conclusion: $f(0) = -2 \times 4 \times 5 \times -1 \times -9 = -360$ , which is negative.
Conclusion: $f(0) = -2 \times 4 \times 5 \times -1 \times -9 = -360$ , which is negative.Since $f(0)$ is negative and there are no zeros between $-4$ and $1$ , $f(x)$ is always negative on the interval -4 < x < 1.