© 2025 Perplex Learning Incorporated. All rights reserved.
All content on this website has been developed independently from and is not endorsed by the International Baccalaureate Organization. International Baccalaureate and IB are registered trademarks owned by the International Baccalaureate Organization.
Discussion
Considering that
describe the graph of f(∣x∣) in terms of transformations of the graph of f(x) on the pieces of its domain.
On x≥0 we have
so for x≥0 the graph of y=f(∣x∣) coincides with the graph of y=f(x) (no transformation).
On x<0 we write
Since −x>0 when x<0, the values are taken from the right-hand branch of y=f(x). Hence for x<0 the graph of y=f(∣x∣) is obtained by reflecting the part of y=f(x) with x≥0 in the y-axis.
In summary:
• For x≥0, plot y=f(x) unchanged.
• For x<0, take the portion of y=f(x) on x≥0 and reflect it across the y-axis.
Graphing f(|x|)
To obtain the graph of f(∣x∣), you can think of it as:
For x≥0: ∣x∣=x, so f(∣x∣)=f(x).
For x<0: ∣x∣=−x, so f(∣x∣)=f(−x).
Practically, you take the portion of y=f(x) for x≥0 and reflect it across the y-axis to fill in the x<0 side. The right side of the original graph becomes the entire graph of f(∣x∣).
Powered by Desmos
Discussion
Describe the graph of ∣f(x)∣ in terms of transformations of the graph of f(x).
The graph of y=∣f(x)∣ is obtained from the graph of y=f(x) as follows:
For any x with f(x)≥0, we have ∣f(x)∣=f(x), so the part of the graph of f at or above the x-axis remains unchanged.
For any x with f(x)<0, we have ∣f(x)∣=−f(x), so the part of the graph of f that lay below the x-axis is reflected in the x-axis.
In other words, take the original curve y=f(x), leave everything on or above the x-axis as is, and reflect every piece below the x-axis upward.
Graphing |f(x)|
To obtain the graph of ∣f(x)∣, take the graph of f(x) and:
Leave all points where f(x)≥0 as they are, since ∣f(x)∣=f(x) in that region.
Reflect any parts of the graph where f(x)<0 above the x-axis, because ∣f(x)∣=−f(x) whenever f(x) is negative.
Effectively, every negative y-value becomes positive, mirroring the portion of the curve below the x-axis to above it.
Powered by Desmos
Notice the "sharp corners" where the function touches the x-axis.
Discussion
Consider the graph of g(x)=[f(x)]2.
What happens to points where f(x)<0?
Hence, minima are guaranteed on the graph of g when f(x)=b. Find b.
Part (a) (i)
For any x with f(x)<0, write f(x)=−h where h>0. Then
so the point (x,−h) on the graph of f goes to (x,h2) on the graph of g. In other words, every portion of the original graph lying below the x-axis is sent to a corresponding point above the axis (with its y-value squared).
Thus, as with f(∣x∣), no values of g are negative.
Part (a) (ii)
We have from part (i) that
for all x, so the smallest possible value of g(x) is 0. This occurs exactly when
Therefore the guaranteed minima on the graph of g happen when f(x)=b with
Graphing [f(x)]^2
To sketch [f(x)]2 from f(x):
Square all y-values: for each x, the new y-value is (f(x))2.
Because y2≥0, the entire graph of [f(x)]2 is on or above the x-axis.
Points where f(x)=0 remain on the x-axis.
Any minima below the x-axis become maxima above the x-axis.
Any maxima below the x-axis become minima above the x-axis.
If ∣f(x)∣<1, squaring makes it closer to 0; if ∣f(x)∣>1, squaring makes it larger.
Negative parts of f(x) become positive when squared, so everything below the x-axis is reflected above it, with the magnitude of y-values adjusted according to the square.
Powered by Desmos
Notice the "rounded corners" where the function touches the x-axis, as opposed to the sharp corners for ∣f(x)∣.
Checkpoint
f(x) is graphed below.
Powered by Desmos
Find the local extrema of [f(x)]2.
Select the correct option
Discussion
Consider g(x)=f(x)1.
If f(x)=c, g(x) is guaranteed to have a vertical asymptote. Find c.
What value does g approach for very large values (positive or negative) of f?
Suppose f has a local minimum at x=m and a local maximum at x=n. Assuming f(m) and f(n) are both nonzero, what can you infer about the features of g at x=m and x=n?
Part (a)
We have g(x)=f(x)1. A vertical asymptote in g occurs exactly where the denominator f(x) is zero. Thus we require
so that g(x) blows up. Hence
Part (b)
As ∣f(x)∣ grows without bound (whether f(x)→+∞ or f(x)→−∞), its reciprocal tends to zero. In symbols:
Hence y=0 is a horizontal asymptote of g.
• Wherever f has a vertical asymptote, f→±∞ nearby, so g→0 there too—producing a “flat” approach to y=0.
• Likewise, on the far tails (x→±∞) if f(x) increases or decreases without bound, g(x) again tends to 0.
Because the numerator of g is the constant 1, every time the denominator blows up, the fraction collapses to zero. Thus the horizontal asymptote is always
Part (c)
Since f(m) and f(n) are nonzero, g(x)=1/f(x) is well defined near m,n.
At x=m, f has a local minimum, so for x near m
and hence
which means
for x close to m. Therefore g has a local maximum at x=m.
At x=n, f has a local maximum, so for x near n
and thus
so
for x close to n. Hence g has a local minimum at x=n.
Answer: g has a local maximum at x=m and a local minimum at x=n.
Graphing 1/f(x)
To sketch f(x)1 from f(x):
At every x-value, take the reciprocal of the original y-value: y→y1.
Points where f(x)=0 become vertical asymptotes, since 01 is undefined.
If ∣f(x)∣ is large, then f(x)1 is small (close to the x-axis). Thus, vertical asymptotes of f(x) become horizontal asymptotes at y=0. Conversely, if ∣f(x)∣ is small, then f(x)1 is large.
Local maxima and minima "flip":
a maximum at (a,b) on f(x) becomes a minimum at (a,b1) on f(x)1;
a minimum at (c,d) becomes a maximum at (c,d1). This happens because reciprocal values invert magnitudes.
Example
Sketch the graph of y=x2+2x−31.
The denominator is zero when x=1 or x=−3, so there will be vertical asymptotes there.
The quadratic x2+2x−3 has a minimum at (−1,−4), so there will me a local maxima at (−1,−41).
Since x2+2x−3 gets larger and larger y as x gets larger, there will be a horizontal asymptote at x=0.
Powered by Desmos
Exercise
It is given that f(x)=−2x2+8x+10.
Determine whether the local extremum of f(x)1 is a maximum or minimum, and find its coordinates.
Find the asymptote(s) of f(x)1.
Select the correct option
Chat