Finding \(a\) When Two Functions Have the Same Minimum Value

Problem Statement:

We are given the functions

\[f(x)=e^x−ax\]

and

\[g(x)=ax−\ln x\]

with \(x>0\) for \(g(x)\)

It is given that \(f(x)\) and \(g(x)\) have the same minimum value.

(1) Find \(a\)

(2) Prove that there exists a line \(y=b\) intersecting both curves \(y=f(x)\) and \(y=g(x)\) at three distinct points whose \(x\)-coordinates form an arithmetic progression

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Answer:

(1)

\(f'(x)=e^x-a\)

\(f′(x)=0\) ⟹ \(0=e^x-a\)

∴ \(x= \ln a\)

The minimum value of \(f\) is:

\[f_{min}=e^{(\ln a)}-a (\ln a)=a-a \ln a\]

\(g'(x)=a-\frac{1}{x}\)

\(g'(x)=0\) ⟹ \(0=a-\frac{1}{x}\)

∴ \(x= \frac{1}{a}\)

The minimum value of \(g\) is:

\[g_{min}=a \cdot(\frac{1}{a})-\ln (\frac{1}{a})=1+\ln a\]

∴ \(a-a \ln a=1+\ln a\)

By inspection, \(a=1\)

(2)

For \(f(x)\), when \(x=\ln a=\ln (1)=0\),

\(f_{min}=a-a \ln a=(1)-(1)\ln (1)=1\)

Since \(f′′(x)=e^x>0\), \(f(x)\) is convex.

For \(g(x)\), when \(x=\frac{1}{a}=\frac{1}{(1)}=1\),

\(g_{min}=1+\ln a=1+\ln (1)=1\)

Since \(g′′(x)=\frac{1}{x^2}>0\), \(g(x)\) is convex.

For \(b>1\), there exists a line \(y=b\) intersects \(f(x)\) in two points: one left of \(x=0 (x_{1})\), one right of \(x=0 (x_{2})\).

∴ \(f(x_{1})=f(x_{2})=b\)

Similarly, for \(b>1\), there exists a line \(y=b\) intersects \(g(x)\) in two points: one left of \(x=1 (x_{3}​)\), one right of \(x=1 (x_{4}​)\).

∴ \(g(x_{3})=g(x_{4})=b\)

But we only have three distinct intersection points. That means \(y=b\) must pass through one intersection point where \(x_{2}=x_{3}\) and \(f(x_{2})=g(x_{3})=b\).

\(f(x_{1})=f(x_{2})=g(x_{3})=g(x_{2})\)

∴ \(e^{x_{1}}−x_{1}=x_{2}-\ln x_{2}\)

\(e^{x_{1}}−x_{1}=e^{(\ln x_{2})}-(\ln x_{2})\)

∴ \(x_{1}=\ln x_{2}\)

\(g(x_{4})=g(x_{3})=f(x_{2})\)

∴ \(x_{4}-\ln x_{4}=e^{x_{2}}−x_{2}\)

\(x_{4}-\ln x_{4}=(e^{x_{2}})−\ln (e^{x_{2}})\)

∴ \(x_{4}=e^{x_{2}}\)

The \(x-\)coordinates of the three distinct intersection points are \(\ln x_{2}, x_{2}\) and \(e^{x_{2}}\)

\(x_{2}-x_{1}\)

\(=x_{2}-\ln x_{2}\)

\(=g(x_{2})\)

\(=g(x_{3})\)

\(=f(x_{2})\)

\(=e^{x_{2}}−x_{2}\)

\(=x_{4}-x_{3}\)

∴ the \(x-\)coordinates of the three distinct intersection points form an arithmetic progression.