A function h(x) increases by 1 over every unit interval in x and h(0) = 0. Which could be a function rule for h(x)? Choices: (A)h(x) = x - 1 (B)h(x) = x (C)h(x) = -x (D)h(x) = -x + 1

Check h(0): Check h(0) for each function to see if it equals 0. For (A) h(x) = x - 1, h(0) = 0 - 1 = -1.Eliminate choices: For (B) h(x) = x, h(0) = 0.Check unit increase: For (C) h(x) = -x, h(0) = -0 = 0.Eliminate choice: For (D) h(x) = -x + 1, h(0) = -0 + 1 = 1.Eliminate choices (A) and (D) because h(0) ≠ 0. Now check if the remaining functions increase by 1 for each unit increase in x.For (B) h(x) = x, if x increases by 1, h(x) increases by 1 since h(x+1) = x+1.For (C) h(x) = -x, if x increases by 1, h(x) decreases by 1 since h(x+1) = -(x+1).Eliminate choice (C) because it decreases instead of increases.

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A function $h(x)$ increases by $1$ over every unit interval in $x$ and $h(0) = 0$ . $\newline$ Which could be a function rule for $h(x)$ ? $\newline$ Choices: $\newline$ (A) $h(x) = x - 1$ $\newline$ (B) $h(x) = x$ $\newline$ (C) $h(x) = -x$ $\newline$ (D) $h(x) = -x + 1$

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Math Problems

Calculus

Intermediate Value Theorem

Full solution

Q. A function

h(x)

increases by

1

over every unit interval in

x

and

h(0) = 0

\newline

Which could be a function rule for

h(x)

\newline

Choices:

\newline

(A)

h(x) = x - 1

\newline

(B)

h(x) = x

\newline

(C)

h(x) = -x

\newline

(D)

h(x) = -x + 1

Check $h(0)$ : Check $h(0)$ for each function to see if it equals $0$ . $\newline$ For (A) $h(x) = x - 1$ , $h(0) = 0 - 1 = -1$ .
Eliminate choices: For (B) $h(x) = x$ , $h(0) = 0$ .
Check unit increase: For $(C)$ $h(x) = -x$ , $h(0) = -0 = 0$ .
Eliminate choice: For (D) $h(x) = -x + 1$ , $h(0) = -0 + 1 = 1$ .
Eliminate choice: For (D) $h(x) = -x + 1$ , $h(0) = -0 + 1 = 1$ .Eliminate choices (A) and (D) because $h(0) \neq 0$ . Now check if the remaining functions increase by $1$ for each unit increase in $x$ .
Eliminate choice: For (D) $h(x) = -x + 1$ , $h(0) = -0 + 1 = 1$ .Eliminate choices (A) and (D) because $h(0) \neq 0$ . Now check if the remaining functions increase by $1$ for each unit increase in $x$ .For (B) $h(x) = x$ , if $x$ increases by $1$ , $h(x)$ increases by $1$ since $h(0) = -0 + 1 = 1$ $0$ .
Eliminate choice: For (D) $h(x) = -x + 1$ , $h(0) = -0 + 1 = 1$ . Eliminate choices (A) and (D) because $h(0) \neq 0$ . Now check if the remaining functions increase by $1$ for each unit increase in $x$ . For (B) $h(x) = x$ , if $x$ increases by $1$ , $h(x)$ increases by $1$ since $h(0) = -0 + 1 = 1$ $0$ . For (C) $h(0) = -0 + 1 = 1$ $1$ , if $x$ increases by $1$ , $h(x)$ decreases by $1$ since $h(0) = -0 + 1 = 1$ $6$ .
Eliminate choice: For (D) $h(x) = -x + 1$ , $h(0) = -0 + 1 = 1$ .Eliminate choices (A) and (D) because $h(0) \neq 0$ . Now check if the remaining functions increase by $1$ for each unit increase in $x$ .For (B) $h(x) = x$ , if $x$ increases by $1$ , $h(x)$ increases by $1$ since $h(0) = -0 + 1 = 1$ $0$ .For (C) $h(0) = -0 + 1 = 1$ $1$ , if $x$ increases by $1$ , $h(x)$ decreases by $1$ since $h(0) = -0 + 1 = 1$ $6$ .Eliminate choice (C) because it decreases instead of increases.