f'(x) = 4x - k/x^2
f"(x) = 4 + 2k/x^3
So you want
4 - 2k = 0
see the graph at
https://www.wolframalpha.com/input/?i=2x%5E2+%2B+2%2Fx+for+-3%3Cx%3C0
For what value of k does the graph of f(x) = 2x^2+k/x
has a point of inflection at x = −1.
2 answers
2