The function y=4/3x^3+ax^2+bx+2/3 has 2 turning points, one of which is at (1,1). Determine the values of a and b. Could someone kindly explain the steps please :)

Question

Could someone explain the steps needed to determine a and b? I know it must be simple (similar to simultaneous equations), but I'm not very good at math :(

Answer 1

(1,1) is a point on the curve, which tells us that we can plug x=1 and y=1 into the function:

1=4/3+a+b+2/3=2+a+b, So a+b=-1. If we find a then we can find b. b=-a-1.

At a turning-point the gradient is zero, so we need to differentiate the function:

dy/dx=4x²+2ax+b=0. We are told that (1,1) is a turning-point, so x=1 is a zero of this quadratic derivative. Therefore a factor is x-1. The other factor is q(x-p), where p and q are constants we need to find or relate to a and b. Therefore:

4x²+2ax+b=q(x-1)(x-p)=q(x²-(p+1)x+p)=qx²-q(p+1)x+pq=0.

Equating coefficients, we can see that q=4 because we have 4x²;

We also know that b=pq=4p (constant term);

And the x term tells us that 4(p+1)=-2a, that is, 4p+4=-2a, that is, b+4=-2a.

But b=-a-1 so, -a-1+4=-2a, a=-3, b=-a-1=3-1=2. 

CHECK

y=4x³/3-3x²+2x+2/3, by putting in a=-3 and b=2, our proposed solution.

Let's see what happens if we plug in x=1:

y=4/3-3+2+2/3=1, which proves that (1,1) lies on the curve. 

Now differentiate:

4x²-6x+2=0 at a turning-point. (Incidentally, 4x²-6x+2 factorises: 4(x-1)(x-½) or 2(x-1)(2x-1). So p=½.) We were told that x=1 at a turning-point, so plug in x=1:

4-6+2=0 is true so x=1 is indeed a turning-point.

This confirms the result a=-3 and b=2.

Comments

Thank you very much Sir!

Is this a commonly used method to determine a and b or are there other ways as well?

I recall my lecturer doing it differently however I did not completely understand. He solved it like a simultaneous equation. Can that be done as well?

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Yes, you can solve it using the method of simultaneous equations. I used substitution. Either way would work. There is no particular way to solve a problem. The important thing is to understand and feel comfortable with whatever method you use, preferably the one that seems easier. When I saw this problem I realised that it involved differentiation because it mentioned turning-points, and I also realised that knowing a point on a curve helps us to find the coefficients. Anyone tackling the problem would have to use the same reasoning. The system of equations would probably be:

a+b=-1 and 2a+b=-4

Solving problems in maths is like solving any problem. You look for one clue, and that leads you to another, until you finally put the pieces together and solve the problem. Always step back from the problem first and ask yourself what tools you will need to tackle it. Sometimes a picture or diagram helps.

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I understand now and totally agree with what you say about solving math problems. One little clue can help work out the rest. Thank you very much for your time and help! I really appreciate it   Learning online is no easy task