33 35 37 34 note that g 86 g86 and g x0 gx0 since g g

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33 35 37 34 Note that g (8)=6 g′(8)=6 and g ′′ (x)<0 g″(x)<0. Since g g is concave down, the rate of change of g g is decreasing. This means that ΔgΔx <6 ΔgΔx<6 when x>8 x>8. Therefore: g(10)−g(8)10−8<6⟹g(10)−g(8)2<6⟹g(10)−g(8)<12⟹g(10)< 12+g(8)⟹g(10)<12+22. g(10)−g(8)10−8<6 g(10)−g(8)2<6 g(10)−g(8)<12 g(10)<12+g(8) g(10)<12+22. Therefore g(10)<34 g(10)<34. g(10)=33 g(10)=33 is the only choice.
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Question 7 2 / 2 pts The following figure shows the graph of y=f(x) y=f(x). At one of the labeled points, dydx dydx and d 2 ydx 2 d2ydx2 are both negative. Which point is it? I. IV. Correct! III. II. V.
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Recall the Leibniz notation: dydx =f (x) dydx=f′(x) and d 2 ydx 2 =f ′′ (x) d2ydx2=f″(x). We want a point where both quantities are negative; a location where f(x) f(x) is decreasing and concave down. III. is the only choice. Question 8 2 / 2 pts The following figure shows the graph of y=f(x) y=f(x). Give the signs of the first and second derivatives. Each derivative is either positive everywhere, negative everywhere or zero everywhere. Correct! f (x) f′(x) is negative; f ′′ (x) f″(x) is negative. f (x) f′(x) is zero; f ′′ (x) f″(x) is negative. f (x) f′(x) is negative; f ′′ (x) f″(x) is positive.
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f (x) f′(x) is positive; f ′′ (x) f″(x) is negative. f (x) f′(x) is negative; f ′′ (x) f″(x) is zero. f(x) f(x) is decreasing and concave down everywhere, so f (x)<0 f′ (x)<0 and f ′′ (x)<0 f″(x)<0. Question 9 0 / 2 pts The table below shows y y as a function of x x, so that y=f(x) y=f(x). According to the data in the table, is the derivative of f(x) f(x) negative or positive? Is the second derivative negative, positive, or zero? x 0 5 10 15 20 25 30 35 y=f(x ) 20 14 9 5 2 0 -1 -1.5 The first derivative is positive; the second derivative is negative. The first derivative is positive; the second derivative is positive. The first derivative is negative; the second derivative is zero. You Answered The first derivative is negative; the second derivative is negative. Correct Answer
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The first derivative is negative; the second derivative is positive. f(x) f(x) is decreasing, so f (x)<0 f′(x)<0. Notice that the size of the decrease is getting smaller as x x gets larger. That is f (x) f′(x) is getting less negative. This means the rate of change f (x) f′(x) is increasing, so f(x) f(x) is concave up and f ′′ (x)>0 f″(x)>0. This may be more clear by looking at a plot plot of the points in the table Question 10 0 / 2 pts Let S(t) S(t) be the total capacity of installed solar power in the US (in TeraWatts), where t t is the year. Choose the best interpretation of the statement: For t≥2009 t≥2009, S (t)>0 S′(t)>0 and S ′′ (t)>0 S″(t)>0.
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Note: This example is not necessarily based on real data. Use only the information given in the problem. Since 2009, the installed capacity of solar power has been increasing, but the rate of increase has been declining.
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