f(x)=(5x+1)(4x-8)(x+6) has zeros at x=-6,x=-(1)/(5), and x=2. What is the sign of f on the interval -(1)/(5) < x < 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 Zeros: Since f(x) has zeros at x=-6, x=-(1)/(5), and x=2, we can test the sign of f(x) by picking a test point from the interval -(1)/(5) < x < 2.Select Test Point: Let's pick x=0 as our test point since it's easy to work with and it's within the interval.Evaluate f(x): Plug x=0 into f(x) to see the sign: f(0)=(5*0+1)(4*0-8)(0+6) = (1)(-8)(6).Calculate Sign: Calculate the sign of f(0): (1)(-8)(6) = -48, which is negative.Final Conclusion: Since f(0) is negative and there are no zeros between -(1)/(5) and 2, f(x) must be negative for all x in the interval -(1)/(5) < x < 2.

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f(x)=(5x+1)(4x-8)(x+6) has zeros at
x=-6,x=-(1)/(5), and
x=2.
What is the sign of
f on the interval
-(1)/(5) < x < 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.

$f(x)=(5 x+1)(4 x-8)(x+6)$ has zeros at $x=-6, x=-\frac{1}{5}$ , and $x=2$ . $\newline$ What is the sign of $f$ on the interval \( -\frac{1}{5}

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Full solution

f(x)=(5 x+1)(4 x-8)(x+6)

has zeros at

x=-6, x=-\frac{1}{5}

, and

x=2

\newline

What is the sign of

f

on the interval

-\frac{1}{5}<x<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 Zeros: Since $f(x)$ has zeros at $x=-6$ , $x=-\frac{1}{5}$ , and $x=2$ , we can test the sign of $f(x)$ by picking a test point from the interval -\frac{1}{5} < x < 2.
Select Test Point: Let's pick $x=0$ as our test point since it's easy to work with and it's within the interval.
Evaluate $f(x)$ : Plug $x=0$ into $f(x)$ to see the sign: $f(0)=(5\cdot0+1)(4\cdot0-8)(0+6) = (1)(-8)(6)$ .
Calculate Sign: Calculate the sign of $f(0)$ : $(1)(-8)(6) = -48$ , which is negative.
Final Conclusion: Since $f(0)$ is negative and there are no zeros between $-\frac{1}{5}$ and $2$ , $f(x)$ must be negative for all $x$ in the interval -\frac{1}{5} < x < 2.