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Concept 10 / 10Sets, Relations and Functions

Transformations of Graphs: Shifts, Stretches, Reflections and Modulus

Obtaining graphs of f(x + a), f(x) + a, f(ax), a f(x), f(-x), -f(x), |f(x)|, f(|x|), [f(x)] and f^-1(x) from the graph of f, the order in which to apply several transformations, and using transformed graphs to count solutions of equations.

Figure 1
Many JEE questions give a function and ask about the number of solutions of an equation, the range, or points of non-differentiability. The fastest route is usually to sketch the graph by transforming a known graph step by step. Each basic transformation has a simple geometric meaning. Translations. y=f(x)+ay = f(x) + a: shift the graph up by aa (down if a<0a < 0). y=f(x+a)y = f(x + a): shift the graph left by aa (right if a<0a < 0). The direction feels backwards: the point that was at x=0x = 0 now appears at x=−ax = -a, because f(x+a)f(x + a) at x=−ax = -a equals f(0)f(0). Stretches. y=af(x)y = af(x) with a>0a > 0: stretch vertically by the factor aa (compress if a<1a < 1). Zeros stay fixed. y=f(ax)y = f(ax) with a>0a > 0: compress horizontally by the factor aa (stretch if a<1a < 1). Points on the yy-axis stay fixed. This is why sin⁡2x\sin 2x has half the period of sin⁡x\sin x. Reflections. y=−f(x)y = -f(x): reflect in the xx-axis. y=f(−x)y = f(-x): reflect in the yy-axis. y=f−1(x)y = f^{-1}(x): reflect in the line y=xy = x (when ff is invertible). Modulus and greatest integer transformations, the ones most tested. y=∣f(x)∣y = |f(x)|: keep the parts of the graph above the xx-axis, and reflect the parts below it upwards. The graph is never below the axis. New corners appear where ff crosses the axis (these are the points of non-differentiability of ∣f∣|f|). y=f(∣x∣)y = f(|x|): discard the part of the graph for x<0x < 0, and replace it with the mirror image of the part for x≥0x \ge 0 in the yy-axis. The result is always an even function. A corner may appear at x=0x = 0 if f′(0)≠0f'(0) \neq 0. ∣y∣=f(x)|y| = f(x): keep the part where f(x)≥0f(x) \ge 0 and add its reflection in the xx-axis; discard the part where f<0f < 0. This is not a function in general (it fails the vertical line test). y=[f(x)]y = [f(x)]: draw horizontal lines y=ny = n for integers nn; on each interval where n≤f(x)<n+1n \le f(x) < n + 1 the graph is the segment at height nn. Jumps occur where ff crosses an integer. y={f(x)}=f(x)−[f(x)]y = \{f(x)\} = f(x) - [f(x)]: cut the graph into the strips between consecutive integer heights and slide every piece down into the band 0≤y<10 \le y < 1. The figure shows ∣f(x)∣|f(x)| and f(∣x∣)f(|x|) for a cubic. Order of transformations. When several are combined, work from the inside of the expression outwards, as the composition dictates. For y=2∣sin⁡(x−π3)∣+1y = 2|\sin(x - \frac{\pi}{3})| + 1: start with sin⁡x\sin x, shift right by π3\frac{\pi}{3}, apply modulus, stretch vertically by 22, then shift up by 11. Changing the order generally gives a different graph: ∣f(x)∣+1|f(x)| + 1 is not ∣f(x)+1∣|f(x) + 1|. Counting solutions. The equation f(x)=g(x)f(x) = g(x) has as many real solutions as the graphs of ff and gg have intersection points. For example, the number of solutions of ∣x2−4∣=k|x^{2} - 4| = k is found by drawing the W-shaped graph of ∣x2−4∣|x^{2} - 4| (with a local maximum 44 at x=0x = 0) and a horizontal line: 00 solutions if k<0k < 0, 22 if k=0k = 0, 44 if 0<k<40 < k < 4, 33 if k=4k = 4 and 22 if k>4k > 4. Key Takeaways: - f(x)+af(x) + a: up by aa; f(x+a)f(x + a): left by aa. - af(x)af(x): vertical stretch; f(ax)f(ax): horizontal compression by aa. - −f(x)-f(x): reflect in the xx-axis; f(−x)f(-x): in the yy-axis; f−1f^{-1}: in y=xy = x. - ∣f(x)∣|f(x)|: flip the negative parts up; f(∣x∣)f(|x|): mirror the right half onto the left. - [f(x)][f(x)]: steps at integer crossings; {f(x)}\{f(x)\}: pieces slid into [0,1)[0, 1). - Apply transformations from the inside out. - Pitfall: shifting f(x+a)f(x + a) to the right instead of the left. Brain-Teaser 1: How many solutions does ∣ ∣x∣−2∣=1|\,|x| - 2| = 1 have? Solution: Build the graph in stages: ∣x∣|x| is a V; ∣x∣−2|x| - 2 shifts it down by 22, crossing the axis at x=±2x = \pm 2 and reaching −2-2 at x=0x = 0; the outer modulus flips the part between −2-2 and 22 up, giving a W with value 22 at x=0x = 0 and zeros at ±2\pm 2. The line y=1y = 1 meets this W four times, at x=±1x = \pm 1 and x=±3x = \pm 3. Algebraically, ∣x∣−2=±1|x| - 2 = \pm 1 gives ∣x∣=3|x| = 3 or ∣x∣=1|x| = 1, the same four solutions. Brain-Teaser 2: Find the points where f(x)=∣x2−3∣x∣+2∣f(x) = |x^{2} - 3|x| + 2| is not differentiable. Solution: Start with g(x)=x2−3x+2=(x−1)(x−2)g(x) = x^{2} - 3x + 2 = (x - 1)(x - 2). Then g(∣x∣)g(|x|) mirrors the right half, and has a corner at x=0x = 0 because g′(0)=−3≠0g'(0) = -3 \neq 0. Its zeros are at x=±1,±2x = \pm 1, \pm 2, where it crosses the axis with non-zero slope. Taking the modulus creates corners at those four crossings. So ff is not differentiable at x=0,±1,±2x = 0, \pm 1, \pm 2: five points. Reading the transformations off the formula identifies all the corners without computing any one-sided derivatives. Brain-Teaser 3: The graph of y=f(x)y = f(x) passes through (2,3)(2, 3). Through which point must the graph of y=2f(3−x)−1y = 2f(3 - x) - 1 pass? Solution: We need 3−x=23 - x = 2, so x=1x = 1, and then y=2×3−1=5y = 2 \times 3 - 1 = 5. The point is (1,5)(1, 5). Thinking of it as transformations: f(3−x)=f(−(x−3))f(3 - x) = f(-(x - 3)) reflects in the yy-axis and then shifts right by 33, taking x=2x = 2 to x=1x = 1; the outer operations stretch vertically by 22 and shift down by 11, taking 33 to 55. The inside expression acts on xx in reverse; the outside acts on yy directly.

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Transformations of Graphs: Shifts, Stretches, Reflections and Modulus — DPP

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