Question

What is the range of the following function and how do u get it? 1/sqrt 1+x

Answer 1

to ge tthe range u must get the graph or find the inverse

Answer 2

is it
\[f(x)=\frac{1}{\sqrt{x+1}}\] ?

Answer 3

it should be clear that \(\sqrt{x+1}\geq0\) so this can never be negative, in other words
\[\frac{1}{\sqrt{x+1}}>0\] for all \(x\)
it can never be zero, because a fraction is only zero if there is a zero in the numerator
so the range is \((0,\infty)\)

Answer 4

i guess one more thing needs to be said: the closer \(x\) gets to \(-1\) the smaller \(x+1\) gets, and therefore the larger \(\frac{1}{\sqrt{x+1}}\) gets
that is why it goes to infinity

Answer 5

thamks

Answer 6

yw

Answer 7

How would you do the following initial value problem d^2y/dt^2 + dy/dt = 2t where t is greater then 0 and y(0) = 0 and dy/dx (0) = 1