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A function $f(x)$ increases by $5$ over every unit interval in $x$ and $f(0) = 0$ . Which could be a function rule for $f(x)$ ? $\newline$ Choices: $\newline$ (A) $f(x) = -\frac{x}{5}$ $\newline$ (B) $f(x) = 5x$ $\newline$ (C) $f(x) = 5^x$ $\newline$ (D) $f(x)$ = $x$ + 5

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Intermediate Value Theorem

Full solution

Q. A function

f(x)

increases by

5

over every unit interval in

x

and

f(0) = 0

. Which could be a function rule for

f(x)

?

\newline

Choices:

\newline

(A)

f(x) = -\frac{x}{5}

\newline

(B)

f(x) = 5x

\newline

(C)

f(x) = 5^x

\newline

(D)

f(x)

=

x

+ 5

Eliminate Incorrect Choices: Since $f(0) = 0$ , we can immediately eliminate choices (C) and (D) because $f(0)$ for both would not be $0$ . $f(0)$ for (C) would be $1$ and for (D) would be $5$ .
Check Choice (A): Now let's check choice (A) $f(x) = -\frac{x}{5}$ . If $x$ increases by $1$ , $f(x)$ should increase by $5$ . So, $f(1)$ should be $5$ . But if we substitute $x = 1$ into choice (A), we get $f(1) = -\frac{1}{5}$ , which is not an increase of $5$ .
Check Choice (B): Let's check choice (B) $f(x) = 5x$ . If $x$ increases by $1$ , $f(x)$ should increase by $5$ . So, $f(1)$ should be $5$ . If we substitute $x = 1$ into choice (B), we get $f(1) = 5 \times 1$ , which is $5$ . This matches the condition that the function increases by $5$ for each unit increase in $x$ .

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