1.2 - Finding ranges with a discriminant

Range

The range of a function is the set of outputs attained at inputs in its domain. A value $y$ is in the range exactly when the horizontal line at height $y$ meets the graph.

Worked example 1: setting up a level equation

Question. Rearrange $y=(x^2+1)/(x^2+4)$ as an equation in $x$, and check $y=1$.

Plan. Multiply by the positive denominator and collect powers of $x$.

At $y=1$, the coefficient of $x^2$ vanishes and the equation becomes $3=0$.

Answer. The level equation is $(y-1)x^2+4y-1=0$; the output $1$ is not attained.

Check. Independently, $f(x)=1-3/(x^2+4)<1$ for every real input, confirming that $1$ is impossible.

Worked example 2: an asymptote that is crossed

Question. Repeat the calculation for $g(x)=(x^2+x+1)/(x^2+4)$.

Plan. Collect the linear term as well, then substitute $y=1$ into the level equation.

At $y=1$, this becomes $-x+3=0$, giving $x=3$. The denominator there is $13\ne0$.

Answer. The level equation is $(y-1)x^2-x+4y-1=0$. The output $1$ is attained at $(3,1)$, even though $y=1$ is a horizontal asymptote.

Check. Substituting into the original function gives $g(3)=(9+3+1)/(9+4)=1$.

Real roots of a quadratic

For real coefficients in $Ax^2+Bx+C=0$, with $A\ne0$, real roots exist if and only if

The quadratic formula is in MF19; this real-root criterion is not separately stated there: learn it.

Worked example 3: a missing endpoint

Question. Find the range of $f(x)=(x^2+1)/(x^2+4)$.

Plan. Use its level equation, then check the exceptional output separately.

For $y\ne1$, the equation $(y-1)x^2+4y-1=0$ has

At $y=1/4$, the equation gives $x=0$, which is allowed. At $y=1$, it gives $3=0$, so this endpoint is excluded. Every value between them gives real inputs with nonzero denominator.

Answer. The range is $[1/4,1)$.

Check. The independent form $f=1-3/(x^2+4)$ has minimum $1-3/4=1/4$ at $0$, and tends to $1$ from below without attaining it.

Worked example 4: two attained endpoints

Question. Find the range of $g(x)=(x^2+x+1)/(x^2+4)$.

Plan. Apply the discriminant to $(y-1)x^2-x+4y-1=0$, checking $y=1$ separately.

For $y\ne1$,

The boundary equation $16y^2-20y+3=0$ has roots

The concave quadratic $\Delta$ is nonnegative between these roots. At either boundary, the repeated input is $x=1/(2(y-1))$, a real allowed input. At $y=1$, the linear equation gives the allowed input $3$.

Answer. The range is $[a,b]$, including both endpoints.

Check. Differentiating the original function gives $g'=(-x^2+6x+4)/(x^2+4)^2$. The stationary inputs $3\pm\sqrt{13}$ give heights $a,b$, respectively, independently checking the range boundaries.

Worked example 5: an output that no finite input gives

Question. Find the range of $f(x)=(x+2)/(x+1)$.

Plan. Solve for the input and test the exceptional output.

For $x\ne-1$,

If $y\ne1$, $x=(2-y)/(y-1)$. This cannot equal $-1$, since that would require $2-y=1-y$. If $y=1$, the equation becomes $0=1$.

Answer. Every real output except $1$ is attained: $\mathbb R\setminus\{1\}$.

Check. The independent form $f=1+1/(x+1)$ gives every height above $1$ on $x>-1$ and every height below $1$ on $x<-1$, but never $1$ itself.

Worked example 6: a hole removes another output

Question. Find the range of $g(x)=(x^2-4)/(x^2-x-2)$.

Plan. Preserve both denominator exclusions, then use the preceding inverse expression.

The output $1$ is still impossible. The removed input $2$ would have output $4/3$. For that output, the equation $(y-1)x=2-y$ gives only $x=2$, so no other input restores it. Every other output except $1$ has an allowed input.

Answer. The range is $\mathbb R\setminus\{1,4/3\}$.

Check. Substitute $y=4/3$ directly into the original cross-multiplied quadratic: it reduces to $(x-2)^2=0$. Its only root is excluded. Thus even a nonnegative discriminant can give no valid point on the original graph.

Watch out

$\Delta\ge0$ certifies real roots of the rearranged equation, not permission to divide by zero in the original function. Keep the domain beside the calculation.