How Does the Natural Exponential Function Influence Trigonometric Outcomes?

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1. Let f(x)=sin(e^x)

a. Find 2 values of x satisfying f(x)=0

b. What is the range of f(x)

c. Find the value(s) of x that maximize f on [3.8,4] (use calculus)



2. y=e^x if and only if x=ln y



3.a. x=-infinity because the limit of e^x as x approaches -infinity is 0. and also, x=infinity because limit e^x as a approaches infinity=infinity

b.The range of f(x) is (0,infinity)

c. the values of x that maximize f on 3.81 and 3.95
 
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Homework Statement


Let f(x)=sin(e^x)

a. Find 2 values of x satisfying f(x)=0

b. What is the range of f(x)

c. Find the value(s) of x that maximize f on [3.8,4] (use calculus)

Homework Equations


y=e^x if and only if x=ln y

The Attempt at a Solution


a. x=-infinity because the limit of e^x as x approaches -infinity is 0. and also, x=infinity because limit e^x as a approaches infinity=infinity

b.The range of f(x) is (0,infinity)

c. the values of x that maximize f on 3.81 and 3.95[/QUOTE]
 
Thanks, micromass.

(a) "infiinity" and "-infinity" are NOT "values of x" so I doubt those are acceptable answers. sin(y)= 0 for y any multiple of \pi. You just need to find x such that e^x= \pi and, say, e^x= 2\pi.

(b) Why does the range not include negative numbers? sin(3\pi/2)= -1 and there certainly exist x such that e^x= 3\pi/2. And how could sin(e^x) be larger than 1?

(c) I get more than just two points.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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