Lagrange's theorem for a cubic function

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Let $f(x)=x^3-6x^2+9x$. (a) Verify that the hypotheses of Lagrange’s (mean value) theorem hold on the intervals $[0,2]$ and $[4,6]$. (b) Find the points $c$ predicted by the theorem on each interval.

Solution

(a) Hypotheses. $f$ is a polynomial, hence continuous on every closed interval and differentiable on every open interval: the hypotheses of Lagrange’s theorem hold on both $[0,2]$ and $[4,6]$.

We compute the derivative: $f'(x)=3x^2-12x+9.$

(b) Interval $[0,2]$. The average rate of change is $\frac{f(2)-f(0)}{2-0}=\frac{2-0}{2}=1.$ Imposing $f'(c)=1$: $3c^2-12c+9=1 \iff 3c^2-12c+8=0 \iff c=\frac{12\pm\sqrt{144-96}}{6}=2\pm\frac{2\sqrt3}{3}.$ The only solution in $(0,2)$ is $c_1=2-\frac{2\sqrt3}{3}\approx 0.845.$

Interval $[4,6]$. The average rate of change is $\frac{f(6)-f(4)}{6-4}=\frac{54-4}{2}=25.$ Imposing $f'(c)=25$: $3c^2-12c+9=25 \iff 3c^2-12c-16=0 \iff c=\frac{12\pm\sqrt{144+192}}{6}=2\pm\frac{\sqrt{336}}{6}.$ The only solution in $(4,6)$ is $c_2=2+\frac{\sqrt{336}}{6}\approx 5.055.$

$\boxed{c_1=2-\tfrac{2\sqrt3}{3}\approx 0.845,\qquad c_2=2+\tfrac{\sqrt{336}}{6}\approx 5.055}$