Find Where Increasing/Decreasing f(x)=x-1

$f (x) = x - 1$

Find the derivative.

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By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d x} [- 1]$

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$1 + 0$

Add $1$ and $0$ .

$1$

Set the derivative equal to $0$ .

$1 = 0$

Since $1 \neq 0$ , there are no solutions.

No solution

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

There are no values of $x$ in the domain of the original problem where the derivative of the function is $0$ or undefined.

No critical points found

No points make the derivative $f' (x) = 1$ equal to $0$ or undefined. The interval to check if $f (x) = x - 1$ is increasing or decreasing is $(- \infty, \infty)$ .

$(- \infty, \infty)$

Substitute any number, such as $1$ , from the interval $(- \infty, \infty)$ in the derivative $f' (x) = 1$ to check if the result is negative or positive. If the result is negative, the graph is decreasing on the interval $(- \infty, \infty)$ . If the result is positive, the graph is increasing on the interval $(- \infty, \infty)$ .

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Replace the variable $x$ with $1$ in the expression.

$f' (1) = 1$

The final answer is $1$ .

$1$

The result of substituting $1$ into $f' (x) = 1$ is $1$ , which is positive, so the graph is increasing on the interval $(- \infty, \infty)$ .

Increasing on $(- \infty, \infty)$ since $1 > 0$

Increasing over the interval $(- \infty, \infty)$ means that the function is always increasing.

Always Increasing

Find Where Increasing/Decreasing f(x)=x-1

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