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Worked solution

Find ranges without losing endpoint conditions

Determine the range of each of the following functions. (a) f:x↦ln⁡x−xf:x\mapsto\ln x-x, x∈Rx\in\mathbb R, 1<x≤e21<x\le e^2. (b) g:x↦(x+2)2+exg:x\mapsto(x+2)^2+e^x, x∈Rx\in\mathbb R, x>−4x>-4.

CJC · Functions · Example 3

(a) Read an open maximum limit and a reached minimum

Answer

Rf=[2−e2,−1)R_f=[2-e^2,-1); the Teacher prints [−5.39,−1)[-5.39,-1) with a rounded lower endpoint.

The Teacher graph marks an open point (1,−1)(1,-1) and a closed point (e2,−5.39)(e^2,-5.39). Independently f ′(x)=1/x−1<0f\,\prime(x)=1/x-1<0 for x>1, so the function decreases throughout its printed domain. At the included right endpoint f(e2)=2−e2f(e^2)=2-e^2; as x→1+x\to1^+ the output approaches −1 without reaching it. Continuity gives every value between. The rounded −5.39 is not an exact endpoint.

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Decreasing ln x−x for 1<x≤e², open (1,−1), closed(e²,2−e²), source rounded minimum −5.39.
Decreasing ln x−x for 1<x≤e², open (1,−1), closed(e²,2−e²), source rounded minimum −5.39.
Decreasing ln x−x for 1<x≤e², open (1,−1), closed(e²,2−e²), source rounded minimum −5.39.
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(b) Keep the interior minimum

Answer

Rg=[m,∞)R_g=[m,\infty) where m=(α+2)2+eα≈0.1310409107m=(\alpha+2)^2+e^\alpha\approx0.1310409107 and 2(α+2)+eα=02(\alpha+2)+e^\alpha=0. The source prints [0.131,∞)[0.131,\infty).

The Teacher completes the graph with excluded (−4,4.02)(-4,4.02), y-intercept 5 and included minimum approximately (−2.06,0.131)(-2.06,0.131). Independently g ′(x)=2(x+2)+exg\,\prime(x)=2(x+2)+e^x and g ′′(x)=2+ex>0g\,\prime\prime(x)=2+e^x>0, so the derivative is strictly increasing. It is negative at −4 and positive at −2, giving exactly one zero α≈−2.0635040718\alpha\approx-2.0635040718 inside the domain. The curve decreases then increases without bound, so its minimum m is attained. Excluding −4 does not remove that interior minimum. The printed point 4.02 and minimum 0.131 are rounded values.

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Curve (x+2) squared+e^x for x>−4, open boundary (−4,4.02), y-intercept 5 and attained minimum (−2.06,0.131), increasing right branch.
Curve (x+2) squared+e^x for x>−4, open boundary (−4,4.02), y-intercept 5 and attained minimum (−2.06,0.131), increasing right branch.
Curve (x+2) squared+e^x for x>−4, open boundary (−4,4.02), y-intercept 5 and attained minimum (−2.06,0.131), increasing right branch.
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Calculator steps 1–3: enter a conditional graph

The source screen reads ln(X)−X/(1<X and X≤e²): the slash follows the final X, without parentheses around the entire difference. A true Boolean condition evaluates to 1, leaving ln(X)−X unchanged; a false condition evaluates to 0 and suppresses the point through division by zero. Dividing the whole numerator by that same condition would give the same permitted graph, but is not the literal source entry. This is calculator-specific syntax, not division by an interval.

Enter the rule

Press Y= and enter ln X − X in Y₁.

Add a conditional denominator

Append a slash using ÷, then an opening parenthesis. The source screen places the division after the final X: ln(X)−X/(condition). When the condition is true it equals 1 and the original expression is unchanged; when false, division by zero suppresses the point.

Select the interval condition

Press 2nd → MATH for TEST, highlight CONDITIONS and choose 8: <X and X≤. This supplies two comparisons joined by and.

Enter a conditional graph without losing its domain

  1. 1. Enter the original rule

    Press Y= and enter ln(X)−X in Y1.

  2. 2. Append the calculator condition

    Append ÷ and an opening parenthesis. Press 2nd → MATH, select TEST → CONDITIONS and choose row 8: <X and X≤.

  3. 3. Complete the endpoint conditions

    Use lower bound 1 and upper bound e². The literal source division follows the final X, as shown here.

    Y1
    ln(X)-X/(1<X and X≤e²)

This is calculator-specific Boolean syntax. A true condition evaluates to 1; a false condition gives division by zero and suppresses the input. It is not division by an interval. The allowed inputs are 1 < x ≤ e². Keep the source parentheses and endpoint signs.

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Reconstructed Y₁ entry and TEST/CONDITIONS menu, selecting row 8 to build ln(X)−X/(1<X and X≤e²). Original screens retained in evidence.
Reconstructed Y₁ entry and TEST/CONDITIONS menu, selecting row 8 to build ln(X)−X/(1<X and X≤e²). Original screens retained in evidence.
Reconstructed Y₁ entry and TEST/CONDITIONS menu, selecting row 8 to build ln(X)−X/(1<X and X≤e²). Original screens retained in evidence.
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Calculator steps 4–9: enter bounds and trace

Fill the upper bound

For e² use the source sequence 2nd → ÷ (the e constant) → ^ → 2. Close the parenthesis after the upper bound.

Fill the lower bound

Move left until < is highlighted; press 2nd → DEL to insert 1 before it. The completed restriction is (1<X and X≤e²).

Graph the restricted rule

Press GRAPH. The source shows the restricted shape and warns that its endpoints are not easy to read from this screen.

Remove only the temporary restriction

Press Y= and delete the domain condition, including the slash, leaving Y₁=ln(X)−X.

Graph the natural-domain rule

Press GRAPH. The source says “all real values of x”; qualify this: the logarithm still requires x>0, even after the additional interval condition is removed.

Trace the boundary coordinates

Press TRACE → 1 → ENTER, then TRACE → 2nd → ÷ → x² → ENTER. The screens report X=1,Y=−1 and X=7.3890561,Y=−5.389056. The first input is excluded from the original function and the second included; tracing does not change those endpoint statuses.

Trace coordinates and retain endpoint membership

  1. 1. Fill the bounds

    For e², use 2nd → ÷ (e) → ^ → 2 and close the condition. Move left to < and use 2nd → DEL to insert the lower bound 1.

  2. 2. Graph the restricted rule

    Press GRAPH for 1 < x ≤ e². A calculator picture can make endpoint inclusion hard to read.

  3. 3. Inspect the unrestricted rule

    Remove the condition and slash in Y=, keeping ln(X)−X. Press GRAPH; its natural real domain is still x > 0.

  4. 4. Trace both boundary coordinates

    Use TRACE 1 ENTER, then TRACE at e² using the retained key sequence 2nd → ÷ → x² → ENTER.

    Trace both boundary coordinates
    InputSource display
    1−1
    7.3890561−5.389056

Tracing a boundary coordinate does not include an excluded endpoint. In the original restricted graph, x=1 is excluded and x=e² is included. Removing the condition does not make ln(X) defined at nonpositive inputs.

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Reconstructed conditional expression, restricted/unrestricted positive-domain graphs and trace readings(1,−1),(7.3890561,−5.389056). First source endpoint open, second closed.
Reconstructed conditional expression, restricted/unrestricted positive-domain graphs and trace readings(1,−1),(7.3890561,−5.389056). First source endpoint open, second closed.
Reconstructed conditional expression, restricted/unrestricted positive-domain graphs and trace readings(1,−1),(7.3890561,−5.389056). First source endpoint open, second closed.
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Solution notes

Student contains blank graph/range responses; Teacher completes them. Both copies have all nine calculator rows. Exact calculus/range qualifications and the unique minimizer proof are independent supplements.