the graph of y=10e^2k

Question

It is concave up correct?

Does it have a minimum at x=o?

Answer 1

I assume by k you meant x, not some constant. If so,

y=10e^2x
y'=20e^2x
y"=40e^2x

so at any x, y" is +, so the slope is increasing, in Texas we call that concave up.

when is y'=0?

y'=0=20e^x. take the ln of each side.

ln(0)=ln20+x
but geepers, ln(0) does not exist.

Now lets consider the graph. Assume x can be any real number.

if x is <0, then y=10e^-realnumber equals some very small number, and as the realnumber gets more negative, as in -345, then y gets pretty small. Well, we have a min at x=-inf obviously.

Now for x=0. y=10
now as x>0, lets say .000005,
y=10*e^.00001, or 10+small number. So y starts at 10, and gets bigger.

So all this indicates to me that the curve is continuous, and has a min at x=-inf, and is always concave upwards.

Answer 2

ok thank you