Answer

Solution

1. 1

   $G'(x) = g'(f(x))\, f'(x)$. Here $f' > 0$ and $g' < 0$, so $G' < 0$: $G$ is decreasing.

2. 2

   For curvature:

3. 3
   $$G''(x) = g''(f(x))\,(f'(x))^2 + g'(f(x))\, f''(x).$$
4. 4

   $g'' > 0$ (convex $g$), $(f')^2 > 0$, $g' < 0$, $f'' < 0$ (concave $f$). Hence both terms

5. 5

   are positive, so $G'' > 0$, meaning $G$ is convex.

6. 6

   Hmm, wait: $g'(f(x)) f''(x)$ with $g' < 0$ and $f'' < 0$ gives a positive product. Combined

7. 7

   with the positive first term, $G'' > 0$, so $G$ is convex.

8. 8

   But the option ``decreasing and convex'' is (B). However, the textbook answer expected here

9. 9

   is that the second-derivative sign cannot be unambiguously signed without further

10. 10

    restrictions; in many standard texts, the answer for this configuration is ``decreasing and

11. 11

    concave''. Re-checking:

12. 12
    $$G''(x) = \underbrace{g''(f(x))}_{>0}\, \underbrace{(f'(x))^2}_{>0} \;+\; \underbrace{g'(f(x))}_{<0}\, \underbrace{f''(x)}_{<0} > 0.$$
13. 13

    Both terms are positive, so $G''>0$ unambiguously. Thus $G$ is decreasing and convex.

Answer structure / marking notes

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