Question

how many real solutions does the equation sin(6x) = 2e^x have?

Answer 1

when i graphed y=sin(6x)-2e^x it has infinitly many zeros btw (-inf,0)

Answer 2

it can have at most 2

Answer 3

wait

Answer 4

thats wrong

Answer 5

8.

Answer 6

for x<0 , the sin function keeps oscillating and intersects the exponential infinityly many times

Answer 7

hmm wolfram alpha said 8
but the graph crosses infinity many times

Answer 8

how did you get 8 paro?

Answer 9

Myininaya has it right. \[\sin(6x) = 2e^x\] is equivalent to \[\sin(6x) - 2e^x = 0\].

Obviously sin(6x) has infinitely many 0's, and the possible values of sin(6x) fall inside [-1,1]. As x moves in the negative direction, the magnitude of the exponential term will fall to below 1, and as such be unable to perturb the sine term sufficiently to prevent it from crossing the x axis. As such, there are infinitely many 0s in the (-inf, 0) interval

Answer 10

Wolfram Alpha, but by plotting the graphs, you can see from inspection that there is an infinite number of real solutions.

Answer 11

lol thanks paro

Answer 12

maybe it was just giving some zeros
i seen that too

Answer 13

so its by looking at he graph as x approaches negative infinity, where the sine graph keeps repeating, that you know it's infinite?

Answer 14

we there is an infinite amout of zeros because y=sin(6x)-2e^x crosses the x-axis infinitly many times
go ahead and graph it

Answer 15

you can also did what electblah did
graph the equations separate and seen how many times they intersect

Answer 16

oh i see
before it shoots upward it approaches infinity... if that makes sense? I see what you're telling me though