- #1

- 412

- 2

## Homework Statement

For the following function, find a point of maxima and a point of minima, if these exist

[tex]

f(x) = 12x^6 - 4x^4 + 15x^3 - 1

[/tex]

## The Attempt at a Solution

I first started by finding the 1st and 2nd derivatives:

[tex]

f'(x) = 72x^5 - 16x^3 + 45x^2

[/tex]

[tex]

f''(x) = 288x^3 - 48x^2 + 50x

[/tex]

On equating, f'(x) with 0,

[tex]

x^2(18x^3 - 12x + 15) = 0

[/tex]

hence, i get x = 0. But, at x = 0, f''(x) = 0, and f'''(x) = 50 i.e. the first non-zero derivative is an odd derivative i.e. the third. Hence, clearly, the point x=0 is a point of inflection.

Now, i need to solve for

[tex]

18x^3 - 12x + 15 = 0

[/tex]

I have no idea how to solve this. On using Mathematica, I am getting 3 solutions, out of which one is real:

[tex]

x = \frac{1}{3} \left(-\frac{2 2^{2/3}}{\left(45-\sqrt{1897}\right)^{1/3}}-\frac{\left(45-\sqrt{1897}\right)^{1/3}}{2^{2/3}}\right)

[/tex]

This value, upto 10 significant digits is:

[tex]

x = -1.173395267

[/tex]

And indeed, there is a minima at this point, which I am able to see when I plot the curve of the function:

http://img410.imageshack.us/img410/2632/maximaminimapk7.png [Broken]

But, since a CAS or a calculator is not allowed on this assignment, I am just not able to understand how to tackle this problem without one. Even the value I got using Mathematica, is too complex for me to arrive at for this level of the assignment. Plus, we've not been taught the solution for cubic equations.

Any help is appreciated.

Thanks,

rohan

Last edited by a moderator: