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Let 
f(x)=x ln(x).
What is the absolute minimum value of 
f ?
Choose 1 answer:
(A) 0
(B) 
(1)/(e)
(C) 
-(1)/(e)
(D) 
f has no minimum value

Let f(x)=xln(x) f(x)=x \ln (x) .\newlineWhat is the absolute minimum value of f f ?\newlineChoose 11 answer:\newline(A) 00\newline(B) 1e \frac{1}{e} \newline(C) 1e -\frac{1}{e} \newline(D) f f has no minimum value

Full solution

Q. Let f(x)=xln(x) f(x)=x \ln (x) .\newlineWhat is the absolute minimum value of f f ?\newlineChoose 11 answer:\newline(A) 00\newline(B) 1e \frac{1}{e} \newline(C) 1e -\frac{1}{e} \newline(D) f f has no minimum value
  1. Find Critical Points: To find the absolute minimum value of the function f(x)=xln(x)f(x) = x \ln(x), we need to find the critical points of the function by taking the derivative and setting it equal to zero.
  2. Calculate Derivative: The derivative of f(x)f(x) with respect to xx is f(x)f'(x). Using the product rule for differentiation, we get: f(x)=ddx[xln(x)]=ln(x)+x(1x)=ln(x)+1.f'(x) = \frac{d}{dx} [x \ln(x)] = \ln(x) + x \cdot (\frac{1}{x}) = \ln(x) + 1.
  3. Set Derivative Equal to Zero: Now we set the derivative equal to zero to find the critical points: ln(x)+1=0\ln(x) + 1 = 0.
  4. Solve for x: To solve for x, we subtract 11 from both sides:\newlineln(x)=1\ln(x) = -1.
  5. Exponentiate Equation: To get xx by itself, we exponentiate both sides of the equation: eln(x)=e1e^{\ln(x)} = e^{-1}.
  6. Determine Critical Point Type: Since eln(x)e^{\ln(x)} is just xx, we have:\newlinex=e1=1e.x = e^{-1} = \frac{1}{e}.
  7. Substitute xx into Function: Now we need to determine if this critical point is a minimum, maximum, or neither. We can do this by using the second derivative test or by analyzing the behavior of the first derivative around the critical point. Since the function ln(x)\ln(x) is increasing for x > 0, and since the critical point x=1ex = \frac{1}{e} is the only critical point, this suggests that f(x)f(x) has an absolute minimum at x=1ex = \frac{1}{e}.
  8. Calculate Minimum Value: To find the minimum value of f(x)f(x), we substitute x=1ex = \frac{1}{e} into the original function:\newlinef(1e)=(1e)ln(1e)f\left(\frac{1}{e}\right) = \left(\frac{1}{e}\right) \cdot \ln\left(\frac{1}{e}\right).
  9. Final Answer: We know that ln(1e)\ln(\frac{1}{e}) is the same as ln(1)ln(e)\ln(1) - \ln(e), which simplifies to 010 - 1 because ln(1)=0\ln(1) = 0 and ln(e)=1\ln(e) = 1. So, f(1e)=(1e)(1)=1ef(\frac{1}{e}) = (\frac{1}{e}) \cdot (-1) = -\frac{1}{e}.
  10. Final Answer: We know that ln(1e)\ln(\frac{1}{e}) is the same as ln(1)ln(e)\ln(1) - \ln(e), which simplifies to 010 - 1 because ln(1)=0\ln(1) = 0 and ln(e)=1\ln(e) = 1. So, f(1e)=(1e)(1)=1ef(\frac{1}{e}) = (\frac{1}{e}) * (-1) = -\frac{1}{e}. Therefore, the absolute minimum value of the function f(x)=xln(x)f(x) = x \ln(x) is 1e-\frac{1}{e}, which corresponds to answer choice (C)(C).

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