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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

Find a cubic function of the form f(x) = ax^3 + bx^2 + cx which has a local maximum at f(1) = 0 and a local minimum at f(3) = -2.

I'm stumped out of my mind. I'm pretty sure that both b and c must be negative, but beyond that, no matter how I munge and funge that function, I can't get it to behave. Is there some voodoo calculus trick that I'm missing here?

(I've tried solving the coefficients as a system of equations, but that didn't work. Every time I got it down to one variable with substitutions, the variable would cancel out, leaving me with something stupid like -2=-2. But, maybe I did it wrong.)

Thanks guys.

El que pega primero pega dos veces.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,616

f(x) = ax^3 + bx^2 + cx

f'(x) = 3ax^2 + 2bx + c

Maxima and Minima are where slope=0, ie f'(x)=0

Where slope=0:

x=1: 3a + 2b + c = 0

x=3: 27a + 6b + c = 0

Now, we also know that f(1) = 0 and f(3) = -2:

a + b + c = 0

27a + 9b + 3c = -2

So, now it is substitution time

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,616

3 unknowns and 4 equations ...

27a + 6b + c = 0

27a + 9b + 3c = -2

reduces to 3b+2c=-2, and then b=-2/3 - (2/3)c

a + b + c = 0

==> a = -b-c

==> a = -(-2/3 - (2/3)c)-c

==> a = 2/3 + (2/3)c - c

==> a = 2/3 - (1/3)c

3a + 2b + c = 0

==> 3(2/3 - (1/3)c) + 2(-2/3 - (2/3)c) + c = 0

==> (2 - c) + (-4/3 - (4/3)c) + c = 0

==> 2 - c -4/3 - (4/3)c + c = 0

==> 2 -4/3 - (4/3)c = 0

==> 2/3 - (4/3)c = 0

==> 2 - 4c = 0

==> 2 = 4c

==> c = 1/2

Now, I haven't made a mistake (and my mistake rate is about 10%, so it is likely )

b = -2/3 - (2/3)c = -2/3 - (1/3) = -1

a = 2/3 - (1/3)c = 2/3 - (1/6) = 1/2

Hmm, must have made a miscalc ... this seems odd, not working out ...

Maybe we are missing the "d" term: f(x) = ax^3 + bx^2 + cx + d

I will work on it later today, ryos.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,616

OK, I did it all on paper, and you do need the "d" term.

The results are:

a=1/2

b=-3

c=4.5

d=-2

I have to go now, but can do a better posting later.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,616

OK, this is what I had on paper:

27a + 9b + 3c + d = -2

a + b + c + d = 0

Subtract and get: 26a + 8b + 2c = -2

Now you have 3 equations and 3 unknowns:

(1) 3a + 2b + c = 0

(2) 27a + 6b + c = 0

(3) 26a + 8b + 2c = -2

Using (1): 3a + 2b + c = 0 ==> c = -3a -2b

Combine with (2): 27a + 6b + (-3a -2b) = 0

==> 24a + 4b = 0

==> a = -(1/6)b

Substitute into (3) gets:

26a + 8b + 2c = -2

==> (-26/6)b + 8b + 2c = -2

==> (22/6)b + 2c = -2

==> c = -1 - (22/12) b

Putting all that in (1) gets:

3a + 2b + c = 0

==> 3(-1/6) b + 2b + (-1 - (22/12) b) = 0

==> (-3/6 + 2 - 22/12)b = 1

==> b = -3

Knowing b, we can calculate

a = -(1/6) b = 1/2

c = -1 - (22/12) b = 4.5

and then calculate d:

a + b + c + d = 0

0.5 + (-3) + 4.5 + -2 = 0

DONE

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

Thanks, you're a lifesaver. I had forgotten that you need at least as many equations as unknowns, and didn't think to use the derivatives.

El que pega primero pega dos veces.

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