Porisma

Worked solution

Read a derivative from an oblique asymptote and two extrema

The diagram shows y=f(x)y=f(x) with asymptotes x=3x=3 and y=−x+2y=-x+2, y-intercept 7/37/3, minimum (2,1)(2,1) and maximum (4,−3)(4,-3).

Sketch the curve y=f′(x)y=f^{\prime}(x). marks

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Given function with asymptotes x=3 and y=−x+2, y-intercept 7/3, minimum (2,1) on the left branch and maximum (4,−3) on the right. The left branch tends up without bound at 3 from the left; the right starts at negative infinity and rises to its maximum before falling.
Given function with asymptotes x=3 and y=−x+2, y-intercept 7/3, minimum (2,1) on the left branch and maximum (4,−3) on the right. The left branch tends up without bound at 3 from the left; the right starts at negative infinity and rises to its maximum before falling.
Given function with asymptotes x=3 and y=−x+2, y-intercept 7/3, minimum (2,1) on the left branch and maximum (4,−3) on the right. The left branch tends up without bound at 3 from the left; the right starts at negative infinity and rises to its maximum before falling.

[2010/VJC/Promo/6(ii)]

CJC · Review problems · Question 4

Answer

Derivative zeros (2,0),(4,0)(2,0),(4,0); source asymptotes x=3x=3, y=−1y=-1, with both inner ends rising to +∞+\infty.

Use the two stationary points

The minimum gives derivative zero (2,0)(2,0) with a negative-to-positive sign change. The maximum gives (4,0)(4,0) with a positive-to-negative sign change. The derivative is negative for x<2x<2, positive for 2<x<32<x<3 and 3<x<43<x<4, and negative for x>4x>4.

How did this part go?

Retain the source asymptotes and branch directions

The printed solution has vertical asymptote x=3x=3, with both sides tending to +∞+\infty, and horizontal asymptote y=−1y=-1, approached from above on both outer ends. The left derivative branch increases throughout and the right decreases throughout; neither has an additional stationary point. The limiting gradient −1-1 is the smooth behaviour shown near the original oblique asymptote, rather than a derivative limit forced by the asymptote equation alone.

How did this part go?

Compare your sketch

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Derivative sketch with vertical asymptote x=3 and horizontal asymptote y=−1. Both branches lie above −1, rise to positive infinity near x=3 and cross the x-axis at 2 and 4 respectively. The left branch increases and the right decreases.
Derivative sketch with vertical asymptote x=3 and horizontal asymptote y=−1. Both branches lie above −1, rise to positive infinity near x=3 and cross the x-axis at 2 and 4 respectively. The left branch increases and the right decreases.
Derivative sketch with vertical asymptote x=3 and horizontal asymptote y=−1. Both branches lie above −1, rise to positive infinity near x=3 and cross the x-axis at 2 and 4 respectively. The left branch increases and the right decreases.

Common mistakes

The y-coordinate 7/3 belongs to the function, not its derivative.

Read the hint