Question

find the absolute minimum and absolute maximum values of ( f ) on the given interval.
( f(x)=e^{-x}-e^{-3 x}, quad0,1 )
absolute minimum value
absolute maximum value
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Explanation:

Step1: Find the derivative of \(f(x)\)

Using the chain rule, if \(y = e^{u}\), then \(y^\prime=e^{u}\cdot u^\prime\).
For \(f(x)=e^{-x}-e^{-3x}\), \(f^\prime(x)=-e^{-x}+3e^{-3x}\).
Set \(f^\prime(x) = 0\), so \(-e^{-x}+3e^{-3x}=0\).
Multiply through by \(e^{3x}\) (since \(e^{3x}
eq0\) for all real \(x\)) to get \(-e^{2x}+3 = 0\).
Let \(t = e^{x}\), then \(-t^{2}+3=0\), \(t^{2}=3\), \(t = e^{x}=\sqrt{3}\) (since \(t=e^{x}>0\)), and \(x=\ln\sqrt{3}=\frac{1}{2}\ln3\approx0.55\) which is in the interval \([0,1]\).

Step2: Evaluate \(f(x)\) at critical points and endpoints

• When \(x = 0\): \(f(0)=e^{-0}-e^{-3\times0}=1 - 1=0\).
• When \(x=\frac{1}{2}\ln3\): \(f(\frac{1}{2}\ln3)=e^{-\frac{1}{2}\ln3}-e^{-3\times\frac{1}{2}\ln3}\).

Using the property \(a^{\log_a b}=b\) and \(n\ln m=\ln(m^{n})\), \(e^{-\frac{1}{2}\ln3}=\frac{1}{\sqrt{3}}\), \(e^{-\frac{3}{2}\ln3}=\frac{1}{3\sqrt{3}}\), so \(f(\frac{1}{2}\ln3)=\frac{1}{\sqrt{3}}-\frac{1}{3\sqrt{3}}=\frac{3 - 1}{3\sqrt{3}}=\frac{2}{3\sqrt{3}}=\frac{2\sqrt{3}}{9}\approx0.38\).

• When \(x = 1\): \(f(1)=e^{-1}-e^{-3}=\frac{1}{e}-\frac{1}{e^{3}}=\frac{e^{2}-1}{e^{3}}\approx\frac{7.39 - 1}{20.09}\approx0.32\).

Answer:

absolute minimum value \(0\)
absolute maximum value \(\frac{2\sqrt{3}}{9}\)