Show that the graph is convex for all values of ##x##

[chwala]

Homework Statement

Kindly see attached

Relevant Equations

differentiation

Part (a) no problem...chain rule

$\dfrac{dy}{dx}= (2x+3)⋅ e^{x^2+3} =0$

$x=-1.5$

For part b,

We need to determine and check if $\dfrac{d^2y}{dx^2}>0$

...
$\dfrac{d^2y}{dx^2}=e^{x^2+3x} [(2x+3)^2+2)]$

Now any value of $x$ will always give us, $\dfrac{d^2y}{dx^2}>0$

The other way/approach would be to pick any two points on the curve and check whether the straight line joining the two points lies under/over the curve to ascertain convex property.

 

[anuttarasammyak]

Let me follow you.
$$x^2+3x=x(x+3)=(x+3/2)^2-9/4$$
Its minimum is -9/4 at x= -3/2 where $y=e^{-9/4}$ which is also an only minium for y of
$$\lim_{x \rightarrow -\infty}y=\lim_{x \rightarrow +\infty}y=+\infty$$

 

[chwala]

Let me follow you.
$$x^2+3x=x(x+3)=(x+3/2)^2-9/4$$
Its minimum is -9/4 at x= -3/2 where $y=e^{-9/4}$ which is also an only minium for y of
$$\lim_{x \rightarrow -\infty}y=\lim_{x \rightarrow +\infty}y=+\infty$$

Boss,
Are you certain of the minimum value $\left[-\dfrac{9}{4}\right]$? or i am reading the wrong script! Ok seen what you meant. We have to be careful with the english here the minimum is actually the value $[e^{-\frac{9}{4}}]$.

 

[fresh_42]

You can always use WA to check a solution:

You misunderstood what @anuttarasammyak has written.

In order to minimize $e^{x^2+3x}$ we need to minmize $x^2+3x$ which is at $x=-1.5.$
The minimal value of $x^2+3x$ is therefore $-2.25.$
The minimal value of $e^{x^2+3x}$ is thus $1/e^{(9/4)}.$

All this happens at $x=-1.5.$

 

[Mark44]

The minimal value of $x^2+3$ is therefore $-2.25.$

Typo: the minimal value of $x^2 + 3$ is 3. I'm sure you omitted x in the second term.

 

[fresh_42]

Typo: the minimal value of $x^2 + 3$ is 3. I'm sure you omitted x in the second term.

Thank you. I corrected it.