2022 ONE(e)

If $p$ is a positive real constant, prove that $y = e^{px^2}$ does not have any points of inflection.

You must use calculus and show any derivatives that you need to find when solving this problem.

$y=e^{px^2}$
$\dfrac{dy}{dx}=2pxe^{px^2}$
$\dfrac{d^2y}{dx^2}=2px\cdot 2px\cdot e^{px^2}+2p\cdot e^{px^2}$
$=2pe^{px^2}(2px^2+1)$
At a point of inflexion, $\dfrac{d^2y}{dx^2}=0$
$2pe^{px^2}(2px^2+1)=0$
Equation 1
$2pe^{px^2}=0$
$pe^{px^2}=0$
$px^2=\ln 0$
No solution as $\ln 0$ is defined.
OR
$2pe^{px^2}=0$ has no solutions
because $2pe^{px^2}>0$ for all values of $x$
since $e^{px^2}>0$ for all values of $a$ and $p$ is a positive
Equation 2
$2px^2+1=0$
$x^2=-\dfrac{1}{2p}$
$x=\sqrt{-\dfrac{1}{2p}}$
$2px^2+1=0$ has no real solutions because $p$
is a positive real constant, $-\dfrac{1}{2p}$ is negative
and there is not a real solution when
you take the square root of a negative number.

OR
$2px^2+1=0$ has no real solutions because
$2px^2+1=0$ is always greater than zero because
$p$ is a positive real constant and $x^2$ is always
greater than or equal to zero

OR
$2px^2+1=0$ has no real solutions because the
discriminant is less than zero.
$b^2-4ac=0-4(2p)(1)=-8p$
Since $p$ is a positive real constant.

Therefore, there are no solutions to $\dfrac{d^2y}{dx^2}=0$ and
and $y=e^{px^2}$ has no points of inflection.