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

In summary, @anuttarasammyak is saying that we need to minimize x^2+3x and that this is at x=-1.5. The minimal value of x^2+3x is -2.25 and the minimal value of e^{x^2+3x} is 1/e^{(9/4).}
  • #1
chwala
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Homework Statement
Kindly see attached
Relevant Equations
differentiation
1686629535288.png


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.
 
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  • #2
Let me follow you.
[tex]x^2+3x=x(x+3)=(x+3/2)^2-9/4[/tex]
Its minimum is -9/4 at x= -3/2 where ##y=e^{-9/4}## which is also an only minium for y of
[tex]\lim_{x \rightarrow -\infty}y=\lim_{x \rightarrow +\infty}y=+\infty[/tex]
 
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  • #3
anuttarasammyak said:
Let me follow you.
[tex]x^2+3x=x(x+3)=(x+3/2)^2-9/4[/tex]
Its minimum is -9/4 at x= -3/2 where ##y=e^{-9/4}## which is also an only minium for y of
[tex]\lim_{x \rightarrow -\infty}y=\lim_{x \rightarrow +\infty}y=+\infty[/tex]
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 :biggrin: the minimum is actually the value ##[e^{-\frac{9}{4}}]##.
 
  • #4
You can always use WA to check a solution:
1686664386913.png


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.##
 
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  • #5
fresh_42 said:
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.
 
  • #6
Mark44 said:
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.
 

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