Answer :
Substitute [tex]y[/tex] for [tex]x^2[/tex] to find that this function is equivalent to:
[tex]y^3 + ky^2 - 9y - 27[/tex]
We take the derivative with respect to [tex]y[/tex] to find the local minima, which occur at the solutions to:
[tex]3y^2 + 2ky - 9 = 0[/tex]
Using the quadratic formula gives:
[tex]y = \frac{-2k \pm \sqrt{4k^2 + 108}}{6}[/tex]
By the Trivial Inequality, since [tex]y=x^2[/tex], [tex]y \geq 0[/tex], so [tex]y=\frac{-2k + \sqrt{4k^2 + 108}}{6} = \frac{-k + \sqrt{k^2 + 27}}{3}[/tex].
To determine whether this is a local minimum or a local maximum, we compute the second derivative, which is [tex]6y+2k[/tex]. This is positive for all positive values of [tex]k[/tex] (a reasonable assumption, since things get really messy really quickly otherwise), so the function is concave up at these values. Hence, this value is a local minimum.
Since it is the only local minimum, we conclude that it is the global minimum, since as [tex]x[/tex] goes to positive or negative infinity, the [tex]x^6[/tex] term dominates and makes the function become extremely large.
The arithmetic gets pretty messy from here (I used WolframAlpha to ensure that I didn't make any mistakes), but in short, the rest of the problem pretty much devolves to substituting this value for [tex]y[/tex] back in, which gives you [tex]f(x) = \frac{2k^3}{27} + 3k - \frac{2}{27} \sqrt{(k^2 + 27)^3} - 27[/tex] as the minimum.
Now, to show that there is no maximum, note that [tex]\lim_{x \to \infty} f(x) = \lim_{x \to \infty} x^6 = \infty[/tex]. Hence, the range of [tex]f(x)[/tex] is [tex][\frac{2k^3}{27} + 3k - \frac{2}{27} \sqrt{(k^2 + 27)^3} - 27, \infty)[/tex].
[tex]y^3 + ky^2 - 9y - 27[/tex]
We take the derivative with respect to [tex]y[/tex] to find the local minima, which occur at the solutions to:
[tex]3y^2 + 2ky - 9 = 0[/tex]
Using the quadratic formula gives:
[tex]y = \frac{-2k \pm \sqrt{4k^2 + 108}}{6}[/tex]
By the Trivial Inequality, since [tex]y=x^2[/tex], [tex]y \geq 0[/tex], so [tex]y=\frac{-2k + \sqrt{4k^2 + 108}}{6} = \frac{-k + \sqrt{k^2 + 27}}{3}[/tex].
To determine whether this is a local minimum or a local maximum, we compute the second derivative, which is [tex]6y+2k[/tex]. This is positive for all positive values of [tex]k[/tex] (a reasonable assumption, since things get really messy really quickly otherwise), so the function is concave up at these values. Hence, this value is a local minimum.
Since it is the only local minimum, we conclude that it is the global minimum, since as [tex]x[/tex] goes to positive or negative infinity, the [tex]x^6[/tex] term dominates and makes the function become extremely large.
The arithmetic gets pretty messy from here (I used WolframAlpha to ensure that I didn't make any mistakes), but in short, the rest of the problem pretty much devolves to substituting this value for [tex]y[/tex] back in, which gives you [tex]f(x) = \frac{2k^3}{27} + 3k - \frac{2}{27} \sqrt{(k^2 + 27)^3} - 27[/tex] as the minimum.
Now, to show that there is no maximum, note that [tex]\lim_{x \to \infty} f(x) = \lim_{x \to \infty} x^6 = \infty[/tex]. Hence, the range of [tex]f(x)[/tex] is [tex][\frac{2k^3}{27} + 3k - \frac{2}{27} \sqrt{(k^2 + 27)^3} - 27, \infty)[/tex].