Question

@ganeshie8 HELP!!

Answer 1

this looks like a straightforward problem, where are you stuck exactly..

Answer 2

how did u get 1 ?

Answer 3

did you find the derivative and set it equal to 0 ?

Answer 4

\[s(t) = xe^x\]

differentiating position gives you velocity
\[v(t) = s'(t) = (xe^x)' = e^x + xe^x = e^x(1+x)\]

Answer 5

set the velocity to 0 you get

\[e^x(1+x) = 0\]

since e^x is never 0, \(1+x = 0 \implies x = -1\)

Answer 6

which is in the interval [-5, 5]

so you're right, the velocity becomes 0 only one time

Answer 7

yes the first derivative equals 0 at x = 1

Answer 8

x = 1

Answer 9

(x-3) cancels out leaving you with (x+3) in the denominator

Answer 10

set the denominator equal to 0 and solve x for vertical asymptote

Answer 11

x + 3 = 0

x = ?

Answer 12

\[x^2-9 = x^2 - 3^2 = (x-3)(x+3)\]

Answer 13

the numerator factor (x-3) and hte denominator (x-3) kill each other

Answer 14

yw!