show (x^2 – x + 2) is positive for all values of x???

apaul322

11

Attachment not found

can someone pls help me with both these question?

(edited 5 years ago)

Original post by apaul322

can someone pls help me with both these question?

Can't see the attachment, but in order to show that your quadratic is +ve you should complete the square. You should end up with something of the form

(x-1/2)^2 + (some +ve number)

Clearly, the first term is always +ve and the second term is always +ve. Together they give something that's always +ve

Reply 2

monkeyman0121

20

Well for starters try to find the stationary points of that equation by doing dy/dx then doing dy/dx=0. The x-value(s) for that will be the points on the curve where there are stationary points, just sub the x value into the normal equation to find y. Then when you get a stationary point. (There is only one on this question.) Try to find if it is a minimum or maximum point by doing d^2y/dx^2. Put the x-value for the stationary point into that and see if it is less than 0 or more than 0. (Could be equal as well.) If it is more than zero then it is a minimum point. This means it is the lowest y-value on the curve and it is over 0 so therefore the equation can only be positive for all values of x.

Also the attachment does not work.

Reply 3

monkeyman0121

20

Original post by RDKGames

Can't see the attachment, but in order to show that your quadratic is +ve you should complete the square. You should end up with something of the form

(x-1/2)^2 + (some +ve number)

Clearly, the first term is always +ve and the second term is always +ve. Together they give something that's always +ve

Oh wow I totally forgot this method! (Would my method work at all? Just checking since it is rather long and may confuse the op.)

Reply 4

apaul322

OP

11

Original post by RDKGames

Can't see the attachment, but in order to show that your quadratic is +ve you should complete the square. You should end up with something of the form

(x-1/2)^2 + (some +ve number)

Clearly, the first term is always +ve and the second term is always +ve. Together they give something that's always +ve

Thank u soo much i understand it now...can you also help me with another question...Given that y = (x^2 + 2)e^–2x
show that dy/dx < 0 for all values of x.

Original post by monkeyman0121

Oh wow I totally forgot this method! (Would my method work at all? Just checking since it is rather long and may confuse the op.)

Strictly speaking, no because all you're doing is getting a stationary point and showing whether it is a max or a min. Without any more information, this is not sufficient to convince someone that this equation always spits out +ve values.

Original post by apaul322

Thank u soo much i understand it now...can you also help me with another question...Given that y = (x^2 + 2)e^–2x
show that dy/dx < 0 for all values of x.

What did you get for dy/dx? You want to essentially factor out a -1 entirely and show that everything inside the brackets is always +ve. This way, you will be taking the -ve of something that's always -ve, hence ending up with something that's always -ve.

Reply 7

apaul322

OP

11

Original post by RDKGames

What did you get for dy/dx? You want to essentially factor out a -1 entirely and show that everything inside the brackets is always +ve. This way, you will be taking the -ve of something that's always -ve, hence ending up with something that's always -ve.

i got e^-2x(2x-2x^2-4) for dy/dx but i dont know where to go from here

Original post by apaul322

i got e^-2x(2x-2x^2-4) for dy/dx but i dont know where to go from here

What do you know about

e^{-2x}

??

Equally, what can you show about

2x-2x^2-4

? [HINT: You can rewrite this so that it has a factor of

x^2-x+2

]

Reply 9

apaul322

OP

11

Original post by RDKGames

What do you know about

e^{-2x}

??

Equally, what can you show about

2x-2x^2-4

? [HINT: You can rewrite this so that it has a factor of

x^2-x+2

][/Qi

i took the negative 2 from outside the box but i didnt get what it meant by show that dy/dx < 0 for all values of x.

Original post by apaul322

i took the negative 2 from outside the box but i didnt get what it meant by show that dy/dx < 0 for all values of x.

You have to show that your result is always -ve, that's what the question wants from you.

Reply 11

apaul322

OP

11

Original post by RDKGames

You have to show that your result is always -ve, that's what the question wants from you.

how do u show that?

Reply 12

Prasiortle

13

Original post by apaul322

how do u show that?

The expression is (-2)e^(-2x)(x^2-x+2). Now, -2 is negative, e^(-2x) is always positive since e to any power is positive, and you just proved that x^2-x+2 is always positive. Thus dy/dx is always negative * positive * positive = negative.

Reply 13

apaul322

OP

11

Original post by Prasiortle

The expression is (-2)e^(-2x)(x^2-x+2). Now, -2 is negative, e^(-2x) is always positive since e to any power is positive, and you just proved that x^2-x+2 is always positive. Thus dy/dx is always negative * positive * positive = negative.

i understand it now...thankk youu soo muchh

show (x^2 – x + 2) is positive for all values of x???

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