Question

f(x)=eˆ(x/2)-2 ; is the zero points to the function =-1 or 2 ?????

Answer 1

are you asing the value \(x_0\) where\(f(x_0) = 0\) ?

Answer 2

yes

Answer 3

so start with \(f(x)=0\) then youll get,
\[e^{\frac x 2 }-2 = 0\]

did you solve this??

Answer 4

but i think that e^0 =1 so 1-2=-1 or what?

Answer 5

yes i solve it like this

Answer 6

so you end up with -1 . but you should end up with 0. I mean
\[e^{(\text{something} )}-2 =0 \]

you chose that something as 0 but ended up with -1 but it should be 0

Rather than guessing values I recommend following steps
\[\begin{align}
e^{\frac x 2} -2 &=0 \\
e^{\frac x 2} &= 2 \\
\ln\left(e^{\frac x 2}\right)=\ln 2
\end{align}\]

can you continue and solve for x?

Answer 7

yep

Answer 8

is that the only zero point?

Answer 9

looks so to me :)