Porisma

Worked solution

Differentiate the shape with a sloping asymptote

The diagram shows the graph of y=g(x)y=g(x). Its intersections with the axes are (0,1)(0,1), (1,0)(1,0) and (3,0)(3,0). The asymptotes are x=2x=2 and y=−x+2y=-x+2.

Sketch the graph of y=g′(x)y=g^{\prime}(x), making clear the main relevant features.

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Given decreasing g: intercepts (0,1), (1,0), and (3,0), vertical asymptote x=2 and sloping asymptote y=−x+2. The left branch falls to negative infinity at 2 from the left; the right branch falls from positive infinity at 2 from the right.
Given decreasing g: intercepts (0,1), (1,0), and (3,0), vertical asymptote x=2 and sloping asymptote y=−x+2. The left branch falls to negative infinity at 2 from the left; the right branch falls from positive infinity at 2 from the right.
Given decreasing g: intercepts (0,1), (1,0), and (3,0), vertical asymptote x=2 and sloping asymptote y=−x+2. The left branch falls to negative infinity at 2 from the left; the right branch falls from positive infinity at 2 from the right.

[2013/VJC/Promo/10(ii)]

CJC · Discussion · Question 3

Answer

The source sketch of g′g^{\prime} has asymptotes x=2x=2 and y=−1y=-1, with both branches below −1-1 and tending to −∞-\infty at x=2x=2.

Read the decreasing branches

Both branches of the supplied graph decrease throughout their domains. The gradient graph therefore lies below the x-axis, with no stationary-point zeros. As drawn, the slopes become arbitrarily negative on both sides of x=2x=2.

How did this part go?

Retain the printed derivative asymptotes

The printed solution has vertical asymptote x=2x=2 and horizontal asymptote y=−1y=-1. Both branches lie below y=−1y=-1, fall towards −∞-\infty near x=2x=2, and approach −1-1 from below as x→±∞x\to\pm\infty.

The limiting slope −1-1 matches the smooth approach shown to the oblique asymptote y=−x+2y=-x+2. The intercepts (0,1)(0,1), (1,0)(1,0) and (3,0)(3,0) locate the original curve; none requires an intercept of g′g^{\prime}. No derivative intercept or exact formula is specified by the source.

How did this part go?

Compare your sketch

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Derivative sketch with vertical asymptote x=2 and horizontal asymptote y=−1. Both branches are below −1 and tend to negative infinity near x=2, approaching −1 from below at the far left and far right.
Derivative sketch with vertical asymptote x=2 and horizontal asymptote y=−1. Both branches are below −1 and tend to negative infinity near x=2, approaching −1 from below at the far left and far right.
Derivative sketch with vertical asymptote x=2 and horizontal asymptote y=−1. Both branches are below −1 and tend to negative infinity near x=2, approaching −1 from below at the far left and far right.

Common mistakes

Do not use the original curve’s zeros as zeros of its derivative.

Read the hint