Continuous and smooth junction of a parabola and an exponential

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Consider the function $f(x)=\begin{cases}-x^2-x+1 & x<0\\[4pt] a\,e^{bx} & x\ge0\end{cases}$ (a) Find $a$ and $b$ so that the junction at $x=0$ is continuous and smooth (i.e. differentiable). (b) With those values, study relative and absolute maxima and minima and the limits at the ends of the domain. (c) Find any inflection point. (d) Sketch the graph. (e) Now set $a=1$ and $b=1$ (NOT the values found above): study relative and absolute maxima and minima, and the sup/inf. (f) Find any inflection point. (g) Sketch the graph. (h) Find the inclination in degrees of the tangent line just before and just after the junction point.

Solution

(a) Continuous and smooth junction at $x=0$. Continuity: from the left $f(0^-)=-0-0+1=1$; from the right $f(0^+)=a\,e^{0}=a$. Hence $a=1.$ Differentiability: on the left $f'(x)=-2x-1\Rightarrow f'(0^-)=-1$; on the right $f'(x)=ab\,e^{bx}\Rightarrow f'(0^+)=ab=b$. Imposing $b=-1$: $a=1,\qquad b=-1,\qquad f(x)=\begin{cases}-x^2-x+1 & x<0\\ e^{-x} & x\ge0.\end{cases}$

(b) Maxima/minima and limits. Left branch: parabola with vertex at $x=-\tfrac12$, value $f\!\left(-\tfrac12\right)=-\tfrac14+\tfrac12+1=\tfrac54$; then decreasing. Right branch: $e^{-x}$ decreases from $1$ (at $x=0$) towards $0^+$. Limits: $\displaystyle\lim_{x\to-\infty}f=-\infty$; $\displaystyle\lim_{x\to+\infty}e^{-x}=0^+$. So absolute maximum $f=\tfrac54$ at $x=-\tfrac12$; no minimum (the infimum is $-\infty$).

(c) Inflection. $f''=-2$ on the left, $f''=e^{-x}>0$ on the right; the function is $C^1$ at $x=0$ and the concavity changes from down to up: inflection at $x=0$, $f(0)=1$.

(d) Graph. Descending parabola on the left up to $(0,1)$, joined without a corner to the decreasing exponential $e^{-x}$ tending to $0^+$.

(e) With $a=1,\ b=1$: $f(x)=\begin{cases}-x^2-x+1 & x<0\\ e^{x} & x\ge0\end{cases}$. Continuous at $0$ (both branches equal $1$) but with a corner point (slopes $-1$ and $+1$). Relative maximum $f=\tfrac54$ at $x=-\tfrac12$; relative minimum (corner) at $x=0$ with $f(0)=1$. Right branch $e^{x}\to+\infty$: $\sup f=+\infty$ (no absolute maximum); left branch $\to-\infty$: $\inf f=-\infty$.

(f) Inflection (case e). No inflection: at $x=0$ the function is not differentiable (corner).

(g) Graph (case e). Same parabola on the left, but on the right an increasing exponential $e^{x}$; a corner appears at $(0,1)$.

(h) Inclination of the tangent at the junction. Smooth case: before and after the slope is $-1\Rightarrow \theta=\arctan(-1)=-45^\circ$ (no change). Case $a=b=1$: before $-1\Rightarrow -45^\circ$, after $+1\Rightarrow +45^\circ$. $\boxed{a=1,\ b=-1;\quad f_{max}=\tfrac54;\quad \theta_-=-45^\circ,\ \theta_+=+45^\circ\ (\text{case }a=b=1)}$