Continuous and smooth junction of a parabola and a logarithm

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Consider the function $f(x)=\begin{cases}-x^2-x+1 & x<2\\[4pt] a\ln(bx) & x\ge2\end{cases}$ (a) Find $a$ and $b$ so that the junction at $x=2$ is continuous and smooth. (b) With those values, study relative and absolute maxima and minima and the limits at the ends of the domain.

Solution

(a) Continuous and smooth junction at $x=2$. Differentiability: on the left $f'(x)=-2x-1\Rightarrow f'(2^-)=-5$; on the right $f'(x)=\dfrac{a}{x}\Rightarrow f'(2^+)=\dfrac{a}{2}$. Imposing equality: $\frac{a}{2}=-5\Rightarrow a=-10.$ Continuity: from the left $f(2^-)=-4-2+1=-5$; from the right $f(2^+)=a\ln(2b)=-10\ln(2b)$. Imposing $-10\ln(2b)=-5$: $\ln(2b)=\tfrac12\Rightarrow 2b=\sqrt e\Rightarrow b=\frac{\sqrt e}{2}\approx0.8244.$

(b) Maxima/minima and limits. Left branch: parabola with vertex at $x=-\tfrac12$, value $f\!\left(-\tfrac12\right)=\tfrac54$ (maximum); at $x=2$ it equals $-5$. Right branch: $a\ln(bx)=-10\ln(bx)$ is decreasing (coefficient $a<0$) and tends to $-\infty$. Limits: $\displaystyle\lim_{x\to-\infty}f=-\infty$; $\displaystyle\lim_{x\to+\infty}\big(-10\ln(bx)\big)=-\infty$. So absolute maximum $f=\tfrac54$ at $x=-\tfrac12$; no minimum (the infimum is $-\infty$). $\boxed{a=-10,\quad b=\frac{\sqrt e}{2}\approx0.8244,\quad f_{max}=\tfrac54}$