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f(x)=(x-1)(4x+3)(3x-8) has zeros at
x=-(3)/(4),x=1, and
x=(8)/(3).
What is the sign of
f on the interval
-(3)/(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.

$f(x)=(x-1)(4 x+3)(3 x-8)$ has zeros at $x=-\frac{3}{4}, x=1$ , and $x=\frac{8}{3}$ . $\newline$ What is the sign of $f$ on the interval $-\frac{3}{4}<x<1$ ? $\newline$ 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.

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Full solution

Q.

f(x)=(x-1)(4 x+3)(3 x-8)

has zeros at

x=-\frac{3}{4}, x=1

, and

x=\frac{8}{3}

.

\newline

What is the sign of

f

on the interval

-\frac{3}{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.

Check signs interval: Check the signs of each factor in the interval $-\frac{3}{4} < x < 1$ .
Analyze $x-1$ : For $x-1$ , when $x$ is between $-\frac{3}{4}$ and $1$ , $x-1$ is negative because $1$ is greater than any number in that interval.
Analyze $4x+3$ : For $4x+3$ , when $x$ is $-\frac{3}{4}$ , $4x+3$ equals $0$ . If $x$ is greater than $-\frac{3}{4}$ but less than $1$ , $4x+3$ is positive because $4x+3$ $0$ times any number greater than $-\frac{3}{4}$ plus $4x+3$ $2$ is positive.
Analyze $3x-8$ : For $3x-8$ , when $x$ is between $-\frac{3}{4}$ and $1$ , $3x-8$ is negative because $3$ times any number less than $\frac{8}{3}$ minus $8$ is negative.
Multiply signs: Now, multiply the signs of each factor. Negative (from $x-1$ ) times positive (from $4x+3$ ) times negative (from $3x-8$ ) gives a positive result.
Final result: Since the product of the signs of the factors is positive, $f$ is always positive on the interval $-\frac{3}{4} < x < 1$ .

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