## anonymous 5 years ago What is the minimum value of f(x) for x > 0, if f(x) = ((x + 1/x)^6 - (x^6 + 1/(x^6)) - 2)/ ((x + 1/x)^3 + (x^3 + 1/(x^3)))

1. anonymous

Wow, that's messy! Have you graphed it to help you make a guess? With a rational function like that, the first thing I do is check the "endpoints". This function seems to have been constructed so that it has no vertical asymptotes for x>0, so I'm guessing the graph is U shaped. (I'm basing this guess on the fact that the function combines x and 1/x so that it "blows up" near x=0 and x=infinity, the fact that the denominator is always positive, and the fact that the sixth power in the numerator should dominate the third power in the denominator.) If I'm right, you'll want to calculate the first derivative and set it equal to zero. You might find that the "binomial theorem" helps you with some of the polynomial multiplication.

2. anonymous

the best method is however to use the AM GM inequality.... AM>=GM.. equality exists when the two numbers are equal....so here x+1/x >=2...put it

3. anonymous

if you put the -2 into that set of nested parentheses in the numerator - i.e. the equation that you wrote down =$((x+1/x)^6-(x^6+1/x^6+2))/((x+1/x)^3+(x^3+1/x^3))$ quickly makes it evident the problem can be reduced to a difference of squares because $(x^3+1/x^3)^2=(x^6+1/x^6+2)$which means that your nasty-looking rational function actually simplifies to the less-gnarly $(x+1/x)^3-(x^3+1/x^3)$ from there, $f \prime(x)=3(x+1/x)^2(1-1/x^2)-3(x^2-1/x^4)$ which, after a little simplification, becomes $f \prime(x)=3-3/x^2$ setting f' equal to 0 yields x=1. Since this is the only extrema and it lies on your given interval (x>0), you know it's your minimum.