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Describing the exponential function y=2e^0.5x^2 
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#1
Jun1511, 12:49 PM

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1. The problem statement, all variables and given/known data
Use the function y=2e^0.5x^2 to answer the following questions a) state the domain b) Determine the intercepts, if any c) Discuss the symmetry of the graph d) Find any asymptotes e)determine the intervals of increase and decrease f)what is the maxima and/or minima g) where is the curve concave upwards and downwards? h) locate the points of inflection 2. Relevant equations 3. The attempt at a solution im having the most difficult with understanding this function since i have never worked with anything like this but this is what i have so far, i would appreciate help me with the steps i have done wrong or dont know how to complete a) state the domain 22<x<22 b) Determine the intercepts, if any yintercept: y=2 the point is (0.2) xintercept: there are no xintercepts c) Discuss the symmetry of the graph the graph is symmetric with respect to the yaxis, as the original equation is unchanged when x is replaced by x d) Find any asymptotes vertical asymptotes: as x approaches 22, e^x decreases to 0; and as x approaches 22 e^x decreases to 0. Therefore the yaxis is a vertical asymptote. vertical asymptotes: there are no vertical asymptotes as the graph never crosses the x axis e)determine the intervals of increase and decrease y=2e^0.5x^2 dy/dx=2e^0.5x^2 * d/dx (0.5x^2) dy/dx= 3xe^0.5x^2 since the sign of the first derivative is the different then the sign of x dy/dx<0 when x>0, and dy/dx<0 when x>0 ( im pretty sure that ones wrong) f)what is the maxima and/or minima Max: (0,2) Min: 0 (not sure about that) g) where is the curve concave upwards and downwards? dy/dx=3xe^0.5x^2 second derivative: 3x d/dx(e^0.5x^2) + e^0.5x^2 d/dx (3x) 3x * e^0.5x^2 * 2x + e^0.5x^2 * 3 = 3e^0.5x^2 (3x^2 +1) the second derivative is negative meaning the graph is concave down h) locate the points of inflection there are no inflection points (unsure about that aswell) i know this is so long but i would really appreciate help on this problem i would like to learn how to do this right. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
Jun1511, 01:06 PM

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P: 21,284

1) The function is not e^x, which is positive for all real x. e^(22) is larger than zero. e^22 is a very large number. 2) Whether the graph crosses the xaxis has nothing to do with vertical or horizontal asymptotes. 


#3
Jun1511, 02:24 PM

P: 7

thankyou for your help yes it is 2e^(o.5x^2)
and iv looked over your comments and tried this again this is what i came up with still have difficulty on some parts a) state the domain  the domain is all real numbers b) Determine the intercepts, if any  yinter (0,2)  xinter are non c) Discuss the symmetry of the graph  the graph is symmetric with respect to the yaxis, as the original equation is unchanged when x is replaced by x d) Find any asymptotes  im really confused how to do this now?? e)determine the intervals of increase and decrease  y=2e^0.5x^2 dy/dx=2e^0.5x^2 * d/dx (0.5x^2) dy/dx= 2xe^0.5x^2 since the sign of the first derivative is the different (neg) then the sign of the function the graph rises when xE (22,0) and falls when xE(22,0). the turning point occurs when x=0, y=2. f)what is the maxima and/or minima Max: (0,2) g) where is the curve concave upwards and downwards? ???? im having trouble with the second derivative i have: = 2x d/dx (e^0.5x^2) + e^0.5x^2 d/dx(2x) = 2x * (0.5x^2)e + e^(0.5x^2) * 2 (x^3)e + 2e^(0.5x^2) but im not sure if thats right or where to go from there i do know that the second derivative should be negative for all x, becuase the graph is concave downwards h) locate the points of inflection inflection point at (0,2) ??? 


#4
Jun1511, 02:37 PM

Sci Advisor
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Thanks
P: 5,076

Describing the exponential function y=2e^0.5x^2
f:=exp(x^2/2): evalf(subs(x=75,f)); 1221 0.3521840206 10 (this is 0.3521840206e(1221)) evalf(subs(x=500,f)); 54286 0.1547968410 10 (this is 0.1547968410e(54286)) Of course, in something like a spreadsheet the program will print out "0" (or else give an underflow warning), but it is not really zero; it is just smaller then the smallest number that program can handle. Theoretically, f(x) = exp(x^2/2) is defined for ALL real x; there is not upper or lower limit. RGV 


#5
Jun1511, 02:53 PM

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P: 21,284

For horizontal asymptotes, do the y values approach a specific value as x approaches infinity or neg. infinity? y = 2e^(.5x^2) dy/dx consists of three factors: 2, x, and e^(.5x^2). 2 is (obviously) always negative. x can be negative, zero, or positive. e^(.5x^2) is ALWAYS positive. On what interval(s) is dy/dx negative? On what interval(s) is dy/dx positive? Where is dy/dx equal to zero? In your first line, it should be = 2x d/dx (e^0.5x^2) + e^0.5x^2 d/dx(2x) In your second line, things go downhill. The derivative of e^(.5x^2) is xe^(.5x^2). 


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