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Determine intervals where a function is positive or negative

Determining where a function is positive or negative means identifying the intervals in its real domain where f(x)>0f(x)>0 or f(x)<0f(x)<0, using its graph, table, or algebraic form. The analysis relates sign changes to x-intercepts, zeros, and domain restrictions such as vertical asymptotes, while recognizing that a function may touch the x-axis without changing sign; boundary zeros and excluded values are not included in the intervals. Complex-valued functions and abstract domains are outside this scope.

Detailed Explanation: Determine intervals where a function is positive or negative

To determine where a function is positive or negative, find the important xx-values where its sign could change:

  • Zeros: values where f(x)=0f(x)=0
  • Excluded values: values where the function is undefined, such as a vertical asymptote

These values divide the number line into intervals. Test one number from each interval to determine whether f(x)f(x) is positive or negative.

Example

Determine where

f(x)=(x−2)(x+1)x−3f(x)=\frac{(x-2)(x+1)}{x-3}

is positive and negative.

1. Find the zeros and excluded values

The numerator is zero when

x−2=0orx+1=0,x-2=0 \quad\text{or}\quad x+1=0,

so the zeros are

x=2andx=−1.x=2 \quad\text{and}\quad x=-1.

The denominator is zero when

x−3=0,x-3=0,

so x=3x=3 is excluded from the domain.

These values divide the number line into four intervals:

(−∞,−1),(−1,2),(2,3),(3,∞).(-\infty,-1),\quad (-1,2),\quad (2,3),\quad (3,\infty).

2. Test one value in each interval

IntervalTest valueSign of f(x)f(x)
(−∞,−1)(-\infty,-1)x=−2x=-2Negative
(−1,2)(-1,2)x=0x=0Positive
(2,3)(2,3)x=2.5x=2.5Negative
(3,∞)(3,\infty)x=4x=4Positive

For example, at x=0x=0,

f(0)=(0−2)(0+1)0−3=(−)(+)(−)>0.f(0)=\frac{(0-2)(0+1)}{0-3} =\frac{(-)(+)}{(-)}>0.

Thus, f(x)f(x) is positive on the interval containing 00.

3. State the answer

The function is positive where

f(x)>0 on (−1,2)∪(3,∞)\boxed{f(x)>0\text{ on }(-1,2)\cup(3,\infty)}

The function is negative where

f(x)<0 on (−∞,−1)∪(2,3)\boxed{f(x)<0\text{ on }(-\infty,-1)\cup(2,3)}

The endpoints are not included because f(−1)=0f(-1)=0 and f(2)=0f(2)=0, while f(3)f(3) is undefined.

Learn by doing: Determine intervals where a function is positive or negative

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Rational Function Inequalities - Factored Quadratic over Binomial - Sign Chart


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