When comparing the functions [tex]f(x)=x^2-x[/tex] and [tex]g(x)=\log(2x+1)[/tex], on which interval are both functions positive?

A. [tex]$(-\infty, 0)$[/tex]
B. [tex]$(0, 1)$[/tex]
C. [tex]$(1, \infty)$[/tex]
D. [tex]$(-\infty, \infty)$[/tex]



Answer :

Certainly! Let's analyze the functions [tex]\( f(x) = x^2 - x \)[/tex] and [tex]\( g(x) = \log(2x + 1) \)[/tex], and determine on which interval both functions are positive.

Firstly, let's examine [tex]\( f(x) \)[/tex]:

[tex]\[ f(x) = x^2 - x \][/tex]

To find where [tex]\( f(x) \)[/tex] is positive, we need to solve the inequality:

[tex]\[ x^2 - x > 0 \][/tex]
[tex]\[ x(x - 1) > 0 \][/tex]

This quadratic inequality factors into the product of two binomials. The solutions to [tex]\( x(x - 1) = 0 \)[/tex] are [tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex]. To determine where the product is positive, we analyze the intervals defined by these critical points:

- For [tex]\( x < 0 \)[/tex], both [tex]\( x \)[/tex] and [tex]\( x-1 \)[/tex] are negative, so their product is positive.
- For [tex]\( 0 < x < 1 \)[/tex], [tex]\( x \)[/tex] is positive and [tex]\( x-1 \)[/tex] is negative, so the product is negative.
- For [tex]\( x > 1 \)[/tex], both [tex]\( x \)[/tex] and [tex]\( x-1 \)[/tex] are positive, so their product is positive.

Therefore, [tex]\( f(x) > 0 \)[/tex] for [tex]\( x \in (-\infty, 0) \cup (1, \infty) \)[/tex].

Next, let's examine [tex]\( g(x) \)[/tex]:

[tex]\[ g(x) = \log(2x + 1) \][/tex]

To find where [tex]\( g(x) \)[/tex] is positive, we need to solve the inequality:

[tex]\[ \log(2x + 1) > 0 \][/tex]

Since the natural logarithm [tex]\(\log(y)\)[/tex] is positive when [tex]\( y > 1 \)[/tex]:

[tex]\[ 2x + 1 > 1 \][/tex]
[tex]\[ 2x > 0 \][/tex]
[tex]\[ x > 0 \][/tex]

Therefore, [tex]\( g(x) > 0 \)[/tex] for [tex]\( x \in (0, \infty) \)[/tex].

Now let's combine these intervals to determine where both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are positive simultaneously:

- For [tex]\( (-\infty, 0) \)[/tex], [tex]\( g(x) \)[/tex] is not positive because [tex]\( g(x) \)[/tex] is positive only for [tex]\( x > 0 \)[/tex].
- For [tex]\( (0, 1) \)[/tex], [tex]\( f(x) \)[/tex] is not positive because we found [tex]\( f(x) \)[/tex] to be negative in this interval.
- For [tex]\( (1, \infty) \)[/tex], [tex]\( f(x) \)[/tex] is positive and [tex]\( g(x) \)[/tex] is positive.

After thorough analysis, we conclude that there is no common interval where both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are positive.

Thus, the result shows that there is no interval where both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are positive simultaneously.