Porisma

Worked solution

Distinguish vertical and horizontal line tests

State and justify which of the following is a one-to-one function. (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 4

(a) Each attained output identifies one input

Answer

f is one-to-one.

The Teacher states that every horizontal line y=k meets the graph at most once. Independently f ′(x)=1/x−1<0f\,\prime(x)=1/x-1<0 throughout the allowed interval, so strict decrease proves this globally. Keep the open point at 1 and included point at e².

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The restricted decreasing logarithmic curve: every horizontal line meets it at most once.
The restricted decreasing logarithmic curve: every horizontal line meets it at most once.
The restricted decreasing logarithmic curve: every horizontal line meets it at most once.
How did this part go?

(b) Find a repeated output

Answer

g is not one-to-one.

The Teacher adds the line y=2, which meets both sides of the minimum. Independently the minimum is below 2, the left boundary limit is 4+e−4>24+e^{-4}>2, and g(0)=5>2g(0)=5>2. Continuity and strict decrease/increase on the two sides give one input for y=2 on each side, so two distinct allowed inputs share an output.

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Curve (x+2) squared+e^x with horizontal Teacher line y=2, crossing both sides of the minimum. Excluded boundary at −4 and y-intercept 5 retained.
Curve (x+2) squared+e^x with horizontal Teacher line y=2, crossing both sides of the minimum. Excluded boundary at −4 and y-intercept 5 retained.
Curve (x+2) squared+e^x with horizontal Teacher line y=2, crossing both sides of the minimum. Excluded boundary at −4 and y-intercept 5 retained.
How did this part go?

Solution notes

The same curves are deliberately reused for a new ask: existence as a function and existence as an inverse use different line tests. Source outlines refer strict monotonicity to Applications of Differentiation.