f(x)=(5x+3)(x-2)(3x+7)(x+5) has zeros at x=-5,x=-(7)/(3),x=-(3)/(5), and x=2. What is the sign of f on the interval -(7)/(3) < x < -(3)/(5) ? 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 Zeros: f(x) has zeros at x=-5, x=-(7)/(3), x=-(3)/(5), and x=2. To determine the sign of f(x) on the interval -(7)/(3) < x < -(3)/(5), we need to test a value within this interval.Choose Test Value: Choose a test value between -(7)/(3) and -(3)/(5), let's pick x=-1. Substitute x=-1 into f(x) to determine the sign.Substitute and Determine Sign: f(-1) = (5(-1)+3)((-1)-2)(3(-1)+7)((-1)+5) = (-2)(-3)(4)(4) = 96. Since 96 is positive, f(x) is positive for our test value.Continuity and Sign: Since f(x) is a polynomial function, it is continuous and will not change sign between zeros unless it crosses a zero. There are no zeros of f(x) between -(7)/(3) and -(3)/(5), so f(x) must be positive throughout this interval.

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f(x)=(5x+3)(x-2)(3x+7)(x+5) has zeros at
x=-5,x=-(7)/(3),x=-(3)/(5), and
x=2.
What is the sign of
f on the interval
-(7)/(3) < x < -(3)/(5) ?
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.

$f(x)=(5 x+3)(x-2)(3 x+7)(x+5)$ has zeros at $x=-5, x=-\frac{7}{3}, x=-\frac{3}{5}$ , and $x=2$ . $\newline$ What is the sign of $f$ on the interval \( -\frac{7}{3}

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Algebra 1

Domain and range of quadratic functions: equations

Full solution

f(x)=(5 x+3)(x-2)(3 x+7)(x+5)

has zeros at

x=-5, x=-\frac{7}{3}, x=-\frac{3}{5}

, and

x=2

\newline

What is the sign of

f

on the interval

-\frac{7}{3}<x<-\frac{3}{5}

\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 Zeros: $f(x)$ has zeros at $x=-5$ , $x=\frac{-7}{3}$ , $x=\frac{-3}{5}$ , and $x=2$ . To determine the sign of $f(x)$ on the interval \frac{-7}{3} < x < \frac{-3}{5}, we need to test a value within this interval.
Choose Test Value: Choose a test value between $-\frac{7}{3}$ and $-\frac{3}{5}$ , let's pick $x=-1$ . Substitute $x=-1$ into $f(x)$ to determine the sign.
Substitute and Determine Sign: $f(-1) = (5(-1)+3)((-1)-2)(3(-1)+7)((-1)+5) = (-2)(-3)(4)(4) = 96$ . Since $96$ is positive, $f(x)$ is positive for our test value.
Continuity and Sign: Since $f(x)$ is a polynomial function, it is continuous and will not change sign between zeros unless it crosses a zero. There are no zeros of $f(x)$ between $-\frac{7}{3}$ and $-\frac{3}{5}$ , so $f(x)$ must be positive throughout this interval.