For the following function, find a point of maxima and a point of minima, if these exist
f(x) = 12x6 – 4x4 + 15x3 –1.
$\displaystyle f(x) = 12x^6-4x^4+15x^3-1$
First, find the derivative of the function:
$\displaystyle f'(x) = 72x^5-16x^3+45x^2$
Now find the critical points. These occur when the derivative function is equal to zero or when it is undefined. Since the function is never undefined, the critical points occur only when it is equal to 0.
So, find the critical points. Then, create a sign chart. Where the sign changes from positive to negative you have a maximum and where it changes from negative to positive you have a minimum.
Hello,
Firstly, derivate the function.
$\displaystyle f'(x)=72x^5-16x^3+45x^2=x^2(72x^3-16x+45)$
0 annulates the derivative twice. But it's not a maxima or minima. Take a look at "inflexion point"
Now i'm afraid you will have to derivate $\displaystyle 72x^3-16x$ to get its variations of sign.
Then, make a sign table of the derivative of f and find its maxima and minima (it's quite hard to write it down on a computer...)
if you use math and /math in [] and this code you will get the function below.
f(x)=12x^6-4x^4+15x^3-1
I asuume you mean
$\displaystyle f(x)=12x^6-4x^4+15x^3-1$
$\displaystyle \frac{df}{dx}=72x^5-16x^3+45x^2=x^2(72x^3-16x+45)$
now lets focus on
$\displaystyle g(x)=72x^3-16x+45$
well
$\displaystyle g(-1)=-72+16+45=-11 \mbox{ and }g(0)=45$
so we know that it has at least one real root inbetwwen -1 and 0
if it is rational you can use the rational roots theorem, if not try numerical methods.
I hope this will get you started. Good luck.