Finding Local Max Min & Absolute Max Min & Intervals

[rsgunter21]

f(x) = x^4 + 2x^3 + 3

Of course the derivate is:

f '(x) = 4x^3 + 6x^2

Which gives the domain:

{x | x /= 0 , x/= -3/2 }

I can see on my calculator where the points of inflections, local min and max, and intervals. I want to know how to do this without a calculator, as my teacher asks for.

Thanks,
Ryan

 

[galactus]

To find your min. Set your derivative to 0 and solve for x.

Inflection points, use the 2nd derivative.

 

[rsgunter21]

Considering this graph had a maximum, how would I find that?

Thanks,
Ryan

 

[galactus]

Suppose that f is twice differentiable at a stationary point, then:

If \(\displaystyle f''(x_{0})<0, \text{then f has a relative max at x_{0}}\)

If \(\displaystyle f''(x_{0})>0, \text{then f has a relative minimum at x_{0}}\)

f is a polynomial and continuous everyhwere.

\(\displaystyle \lim_{x\to\{+\infty}}{x^{4}+2x^{3}+3}=+\infty\)

\(\displaystyle \lim_{x\to\{-\infty}}{x^{4}+2x^{3}+3}=+\infty\)

So f has a minimum, but no maximum on \(\displaystyle (-\infty,+\infty)\)

f'(x)=0 gives us critical points at x=0 and x=-3/2

f''(x)=0 gives us an inflection at x=-1

But, f''(-3/2)=9. 9 not less than 0. Not a relative max.