Question 4 2 2 pts calculate the limit lim x0 e x 2x

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Question 4 2 / 2 pts Calculate the limit: lim x→0 e x −2x 2 limx→0ex−2x2 0 0 e e Correct! 1 1 2 2 e−2 e−2 lim x→0 e x −2x 2 =e 0 −2(0) 2 =e 0 −0=1 limx→0ex−2x2=e0−2(0)2=e0−0=1 Question 5 2 / 2 pts Evaluate the limit: lim h→0(h+2) 2 −4h limh→0(h+2)2−4h
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−4 −4 2 2 0 0 −2 −2 Correct! 4 4 lim h→0 (h+2) 2 −4h=lim h→0 h 2 +4h+4−4h=lim h→0 h 2 +4hh=lim h→0 h(h+ 4)h=lim h→0 h+4=0+4=4 limh→0(h+2)2−4h=limh→0h2+4h+4−4h=limh→0h2+4h h=limh→0h(h+4)h=limh→0h+4=0+4=4 Question 6 2 / 2 pts A function g g has g(8)=22 g(8)=22, g (8)=6 g′(8)=6, and g ′′ (x)<0 g″(x)<0 for all x x. Which of the following is a possible value for g(10) g(10)? (Assume the function is continuous and the first and second derivatives exist for all x x.) 37
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34 Correct! 33 36 35 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. 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 is zero and d 2 ydx 2 d2ydx2 is positive. Which point is it?
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V. III. IV. Correct! I. II. 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 dydx =0 dydx=0 (the
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tangent to f f is horizontal) and d 2 ydx 2 >0 d2ydx2>0 ( f f is concave up). I. is the only choice. Note that dydx =0 dydx=0 at III. but here d 2 ydx 2 <0 d2ydx2<0 (concave down). 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. f (x) f′(x) is positive; f ′′ (x) f″(x) is negative. Correct! f (x) f′(x) is zero; f ′′ (x) f″(x) is zero. f (x) f′(x) is zero; f ′′ (x) f″(x) is negative.
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f (x) f′(x) is positive; f ′′ (x) f″(x) is zero. f (x) f′(x) is negative; f ′′ (x) f″(x) is zero. The graph of f(x) f(x) is a horizontal line. The tangent is horizontal, so f (x)=0 f′(x)=0, and since f(x) f(x) is linear (the rate of f (x) f′(x) is constant), we have f ′′ (x)=0 f″(x)=0. Question 9 0 / 2 pts The table below shows the Earth's atmospheric pressure p p in kiloPascals (kPa) as a function of the altitude h h, in kilometers (km), so p=f(h) p=f(h). 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? h (altitude in km) 0 2 4 6 8 10 p = f(h) (pressure in kPa) 100 80 6 2 47 35 27 The first derivative is negative; the second derivative is zero.
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