# A2 Maths concavity question

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#1
A question says f(x)=x^4+2x³+3x²-10x+5 is concave for all real x values. FIND THE range of possible values of a.

I've found f''(x) = 12ax²+12x+6
So 2ax²+2x+1≥0
What should i do next? Thank you
0
2 months ago
#2
It helps to post the original question as there is no "a" in your f(x). Im presuming it is the quartic coefficient.
To show a quadratic is positive for all x, you usually complete the square.
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#3
(Original post by mqb2766)
It helps to post the original question as there is no "a" in your f(x). Im presuming it is the quartic coefficient.
To show a quadratic is positive for all x, you usually complete the square.
Oh sorry about that - yea it's the quartic.
I'm not showing its always real tho - i'm finding a?
Thanks
Last edited by Rhys_M; 2 months ago
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2 months ago
#4
Post what you tried when you complete the square.

Im not sure about your always real comment - when you complete the square, what do you show and what does it depend on. If you're unsure, just pick a quadratic example and work it through.
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#5
(Original post by mqb2766)
Post what you tried when you complete the square.

Im not sure about your always real comment - when you complete the square, what do you show and what does it depend on. If you're unsure, just pick a quadratic example and work it through.
I've tried to complete the square below but I can't get anywhere manipulating the answer to find a
Here's the full question and what I've done
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2 months ago
#6
Agree with the expression at the top of the third column,

Tbh, that expression is nearly there, you just need to interpret what your completed square expression means and how it relates to 0 in terms of a by thinking about the quadratic multiplier and the vertical offset. Note you've already assumed a is not 0, when you completed the square. Write this down explicitly as it is (a small) part of the answer.

You can use the expression at the bottom of that column, but the extra derivation isn't really necessary.
Last edited by mqb2766; 2 months ago
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#7
(Original post by mqb2766)
Agree with the expression at the top of the third column,

Tbh, that expression is nearly there, you just need to interpret what your completed square expression means and how it relates to 0 in terms of a by thinking about the quadratic multiplier and the vertical offset. Note you've already assumed a is not 0, when you completed the square. Write this down explicitly as it is (a small) part of the answer.

You can use the expression about the bottom of that column, but the extra derivation isn't really necessary.
So my completed square is right, and I need to state I'm assuming a≠0, but my problem is I do not no how to rearrange for a which I'm assuming I need to do. Can you explain explicitly the maths I need to do?
thanks
0
2 months ago
#8
Its probably easier if you don't rearrange in terms of a, rather interpret the inequality at the top of the third column, as previously suggested. You want to find the values of a for which
2a(x + 1/2a)^2 + 1-1/2a >= 0
for all x. You should be able to interpret the two parts
2a(x + 1/2a)^2
1-1/2a
in terms of what values of a ensure their sum is >= 0?

The two pictures in
https://www.nuffieldfoundation.org/s...%20student.pdf
are what you should be thinking. You want to consider the quadratic coefficient "2a", the vertical offset "1-1/2a", and to show the quadratic is positive for all x, ti.e. the quadratic should not intersect with the x-axis. Considering the turning point x=-1/2a may also help in the reasoning.

If you really can't get started with the inpterpretation, pick some values for a, and plot the quadratics and/or completed square expressions in desmos. What do you notice about it being >= 0 for all x, which values of a are ok, ... You'll learn from doing this.
Last edited by mqb2766; 2 months ago
0
#9
(Original post by mqb2766)
Its probably easier if you don't rearrange in terms of a, rather interpret the inequality at the top of the third column, as previously suggested. You want to find the values of a for which
2a(x + 1/2a)^2 + 1-1/2a >= 0
for all x. You should be able to interpret the two parts
2a(x + 1/2a)^2
1-1/2a
in terms of what values of a ensure their sum is >= 0?

The two pictures in
https://www.nuffieldfoundation.org/s...%20student.pdf
are what you should be thinking. You want to consider the quadratic coefficient "2a", the vertical offset "1-1/2a", and to show the quadratic is positive for all x, ti.e. the quadratic should not intersect with the x-axis. Considering the turning point x=-1/2a may also help in the reasoning.

If you really can't get started with the inpterpretation, pick some values for a, and plot the quadratics and/or completed square expressions in desmos. What do you notice about it being >= 0 for all x, which values of a are ok, ... You'll learn from doing this.
Using desmos, I get a≥1/2 which is the answer but I still don't understand what this means - I want to do the algebraic way as I won't get access to Desmos in my A-level. I don't understand what you mean by 'interpret them' - mathematically how do I get from my completed square to finding a?
0
2 months ago
#10
I know you won;t be able to use desmos in an exam.

Completing the square is an algebraic method. What can you say about the two expression parts either side of a=1/2? What do the quadratics look like. And you can explore in desmos to build understanding.
Last edited by mqb2766; 2 months ago
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#11
(Original post by mqb2766)
I know you won;t be able to use desmos in an exam.

Completing the square is an algebraic method. What can you say about the two expression parts either side of a=1/2? What do the quadratics look like. And you can explore in desmos to build understanding.
Using desmos, Above a=1/2, there are no roots. At a=1/2, there is a root. Below a=1/2 there are 2 roots.
But I don't get what this means in terms of the question and feel like I'm really overcomplicating it.
Algebraically how do I find a=1/2 in the first place? this is all I need to do.
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2 months ago
#12
You're not analyzing whether there are roots, though it is related. You're analyzing whether the quadratic is always positive (strictly being not negative). See #8 (first paragraph) about analyzing the two parts which together give a>=1/2.

The question builds on your understanding of completing a square / function being positive. If you're thinking in terms of roots you probably need to do a few more examples like this to understand the roles of the "2a" and "1-1/2a" terms in the completed square expression.
Last edited by mqb2766; 2 months ago
0
#13
(Original post by mqb2766)
You're not analyzing whether there are roots, though it is related. You're analyzing whether the quadratic is always positive (strictly being not negative). See #8 (first paragraph) about analyzing the two parts which together give a>=1/2.

The question builds on your understanding of completing a square / function being positive. If you're thinking in terms of roots you probably need to do a few more examples like this to understand roles of the "2a" and "1-1/2a" terms in the completed square expression.
I wasn't thinking about roots until you said playing around with desmos will help.
Convex is when f''(x)≥0
So 2ax² +2x+1≥0
I just need to find a - plz can you just explain how to get a 😭😭
Last edited by Rhys_M; 2 months ago
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2 months ago
#14
2a > 0
because the quadratic is "u" rather than "n".
1-1/2a >= 0
because the quadratic (turning point) is above the x-axis. See #8.

Playing around with desmos does not mean looking at the roots. In the previous linked pdf, the completed square was descibed in terms of the turning value ("height") and the convex / concave "a" parameter, so the min/max values which occur at the line of symmetry. The word roots only occurs once in the pdf, right at the end, Sometimes a question requires you to interpret terms, not just blindly do algebra.
Last edited by mqb2766; 2 months ago
0
#15
(Original post by mqb2766)
2a > 0
because the quadratic is "u" rather than "n".
1-1/2a >= 0
because the quadratic (turning point) is above the x-axis. See #8.

Playing around with desmos does not mean looking at the roots. In the previous linked pdf, the completed square was descibed in terms of the turning value ("height") and the convex / concave "a" parameter, so the min/max values which occur at the line of symmetry. The word roots only occurs once in the pdf, right at the end, Sometimes a question requires you to interpret terms, not just blindly do algebra.
1-1/2a≥0
So 1≥1/2a
So 2≥a , so a≤2 ?????
doesn't give the answer a≥1/2 what am I missing
0
2 months ago
#16
Without being funny, just do it correctly. Your answer doesn't satisfy the original inequality, its almost a one liner to solve.

You realize the
1/2a
is
1/(2a)
not
a/2
Last edited by mqb2766; 2 months ago
0
#17
(Original post by mqb2766)
Without being funny, just do it correctly. Your answer doesn't satisfy the original inequality, its almost a one liner to solve.

You realize the
1/2a
is
1/(2a)
not
a/2
Thank you soooo much for all this - you're not being funny it's fine I've just completely lost my mind spending all day on the same f**king question . Sorry lol
0
2 months ago
#18
Hi! I can help to make for you The maths term paper. Working on it at home, because I am a new mam and have a baby! Thanks beforehand, will be glad to be useful. 0
2 months ago
#19
Without wanting to dwell on this too much (completing the square, inequalities, desmos, ...), you might want to make sure you're happy with why you only consider the positive solution for a as there are two intervals
https://www.desmos.com/calculator/pv33o92fen
In terms of the original parameterized quadratic
https://www.desmos.com/calculator/ijrdcptstf

Expecting a solution like a=... wasn't the way to go. You had to argue the quadratic was positive, which you could only do if it (completed square quadratic) was convex (2a>0) and the minimum value was greater than or equal to zero (1-1/2a>=0). That was the understanding you seemed to be missing and analyzzing in desmos (even "solving" inequalities) is quick, easy.and a good use of time, possibly even more so if you've spend a long time trying to bash the problem with algebra.

Note that if you just think about the roots, you may forget to check whether the convex/concave property and incorrectly include extra solutions.
Last edited by mqb2766; 2 months ago
0
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