find the absolute minimum and absolute maximum values of ( f ) on the given interval.\n( f(x)=e^{-x}-e^{-3…

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}\neq0) 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})