what is the minimum value of ( f(x)=xln(x) )?\na -e\nb -1\nc ( -\frac{1}{e} )\nd 0\ne has no minimum value

Explanation:

Step1: Find the derivative of the function

The function is (f(x)=x\ln(x)), and its domain is (x>0). Using the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x), (u^\prime=1), (v=\ln(x)), (v^\prime=\frac{1}{x}). So (f^\prime(x)=\ln(x)+x\times\frac{1}{x}=\ln(x) + 1).

Step2: Find the critical points

Set (f^\prime(x)=0), then (\ln(x)+1 = 0). Solve for (x): (\ln(x)=- 1), and by the property of logarithms (x = e^{-1}=\frac{1}{e}).

Step3: Determine the nature of the critical point

Take the second - derivative. (f^\prime(x)=\ln(x)+1), then (f^{\prime\prime}(x)=\frac{1}{x}). When (x = \frac{1}{e}), (f^{\prime\prime}(\frac{1}{e})=e>0). So (x=\frac{1}{e}) is a point of minimum.

Step4: Calculate the minimum value

Substitute (x = \frac{1}{e}) into (f(x)): (f(\frac{1}{e})=\frac{1}{e}\ln(\frac{1}{e})). Since (\ln(\frac{1}{e})=-1), then (f(\frac{1}{e})=-\frac{1}{e}).

Answer:

C. (-\frac{1}{e})