Calculus: Early Transcendentals, Chapter 4, 4.6, Section 4.6, Problem 10

Let c=2\cdot 10^8 . Then rearranging the function, we get f(x)=1/x^8-c/x^4 = x^{-8}-cx^{-4}=\frac{1-cx^4}{x^8} .
The derivative is found using the power rule to get
f'(x)=-8x^{-9}+4cx^{-5} and in factored form this is
=-\frac{4(2-cx^4)}{x^9}
The second derivative can also easily be found using the power rule to get
f''(x)=72x^{-10}-20x^{-6} and in factored form this is
=\frac{4(18-5cx^4)}{x^10}
The function has x-intercepts at x=+-1/c^{1/4}\approx +- 0.008409 , found by setting the numerator to zero in the initial function. When x=0 , we get vertical asymptotes since the denominator of the function is zero. Also, since the numerator of the function is degree four and the degree of the denominator is eight, there is a horizontal asymptote at y=0 .
By setting the numerator of the first derivative to zero, there are local extrema found at x=+-(2/c)^{1/4} = +-0.01 . The first derivative is decreasing on the intervals (-\infty, -0.01) and (0,0.01) and increasing on the intervals (-0.01, 0) and (0.01, \infty) .
The second derivative then provides the inflection points by setting the numerator to zero, which gives the two inflection points at x=+-(18/{5c})^{1/4}\approx +-0.01158 . The function is then concave down on the intervals (-\infty, -0.01158) and (0.01158, \infty) , and concave up on the intervals (-0.01158, 0) and (0, 0.01158) .
Combined, this information gives the graph: