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Using only the values given in the table for the function, [tex]f(x)[/tex], what is the interval of [tex]x[/tex]-values over which the function is increasing?

\begin{tabular}{|c|c|}
\hline[tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline-6 & 34 \\
\hline-5 & 3 \\
\hline-4 & -10 \\
\hline-3 & -11 \\
\hline-2 & -6 \\
\hline-1 & -1 \\
\hline 0 & -2 \\
\hline 1 & -15 \\
\hline
\end{tabular}

A. [tex]$(-6,-3)$[/tex]

B. [tex]$(-3,-1)$[/tex]

C. [tex]$(-3,0)$[/tex]

D. [tex]$(-6,-5)$[/tex]


Sagot :

To determine where the function [tex]\( f(x) \)[/tex] is increasing, we need to examine the given [tex]\( x \)[/tex] and [tex]\( f(x) \)[/tex] values.

Here is the table we have:

[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -6 & 34 \\ \hline -5 & 3 \\ \hline -4 & -10 \\ \hline -3 & -11 \\ \hline -2 & -6 \\ \hline -1 & -1 \\ \hline 0 & -2 \\ \hline 1 & -15 \\ \hline \end{array} \][/tex]

We look for intervals where [tex]\( f(x) \)[/tex] increases as [tex]\( x \)[/tex] increases.

1. From [tex]\( x = -6 \)[/tex] to [tex]\( x = -5 \)[/tex]; [tex]\( f(-6) = 34 \)[/tex] and [tex]\( f(-5) = 3 \)[/tex]
- [tex]\( f(-6) > f(-5) \)[/tex] (decreasing)

2. From [tex]\( x = -5 \)[/tex] to [tex]\( x = -4 \)[/tex]; [tex]\( f(-5) = 3 \)[/tex] and [tex]\( f(-4) = -10 \)[/tex]
- [tex]\( f(-5) > f(-4) \)[/tex] (decreasing)

3. From [tex]\( x = -4 \)[/tex] to [tex]\( x = -3 \)[/tex]; [tex]\( f(-4) = -10 \)[/tex] and [tex]\( f(-3) = -11 \)[/tex]
- [tex]\( f(-4) > f(-3) \)[/tex] (decreasing)

4. From [tex]\( x = -3 \)[/tex] to [tex]\( x = -2 \)[/tex]; [tex]\( f(-3) = -11 \)[/tex] and [tex]\( f(-2) = -6 \)[/tex]
- [tex]\( f(-3) < f(-2) \)[/tex] (increasing)

5. From [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex]; [tex]\( f(-2) = -6 \)[/tex] and [tex]\( f(-1) = -1 \)[/tex]
- [tex]\( f(-2) < f(-1) \)[/tex] (increasing)

6. From [tex]\( x = 0 \)[/tex]; [tex]\( f(0) = -2 \)[/tex]
- [tex]\( x = -1 \)[/tex] to [tex]\( x = 0; f(-1) = -1 \begins an increase and \(f(0)\ decreases decreasing 7. From \( x = 1) ; f(0) = -1; and \( f(1 = -15, 0) \ commences an increase and oint Checking each \( x \)[/tex]:

- The interval from \( x(-3, -2) \( -3 and \(-2( increasing becauseif (\( f(-3)\ <\ f(-2) )

- The interval \( x(-2, -1) \( from becauseif \( f(-2 incrementing -1)

Thus, the function \( f(x) is only between \( -3 -2 and \( -2 -1 respectively.

So, the correct interval where the function is increasing is:

\(\boxed{-3, -1}).