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Exercise 171Calculus › Single-variable calculus › Differentiation › Optimisation and curve sketchingL1Added 4 Nov 2026

A family of functions: x² + m/x

#parameters#extrema#monotonicity#discussion

Statement

For m∈Rm\in\mathbb{R}, let fm(x)=x2+mxf_m(x)=x^2+\dfrac mx on (0,+∞)(0,+\infty). According to mm, study the variations of fmf_m and find its minimum when it exists.

Hint

fm′(x)=2x−mx2=2x3−mx2f_m'(x)=2x-\dfrac m{x^2}=\dfrac{2x^3-m}{x^2}.

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Solution

Case m≤0m\le0. 2x3−m>02x^3-m>0 for x>0x>0: fmf_m is strictly increasing on (0,∞)(0,\infty). It has no minimum: for m<0m<0, fm(x)→−∞f_m(x)\to-\infty as x→0+x\to0^+; for m=0m=0, f0(x)=x2f_0(x)=x^2 has infimum 00 which is not attained on (0,∞)(0,\infty).

Case m>0m>0. fm′f_m' vanishes at x0=(m2)1/3x_0=\left(\dfrac m2\right)^{1/3}, is negative before and positive after: fmf_m decreases on (0,x0](0,x_0] and increases on [x0,∞)[x_0,\infty). Since m=2x03m=2x_0^3,
fm(x0)=x02+2x03x0=3x02=3(m2)2/3.f_m(x_0)=x_0^2+\frac{2x_0^3}{x_0}=3x_0^2=3\left(\frac m2\right)^{2/3}.
This is the global minimum. Also fm→+∞f_m\to+\infty at both ends.

Application. For m=16m=16: x0=2x_0=2 and the minimum is f(2)=4+8=12=3⋅82/3f(2)=4+8=12=3\cdot8^{2/3}.

Remark. The qualitative behaviour changes at m=0m=0, a bifurcation value: for m>0m>0 the term mx\frac mx creates a minimum, for m≤0m\le0 it does not.