Q. Let f(x)=xln(x).What is the absolute minimum value of f ?Choose 1 answer:(A) 0(B) e1(C) −e1(D) f has no minimum value
Find Critical Points: To find the absolute minimum value of the function f(x)=xln(x), we need to find the critical points of the function by taking the derivative and setting it equal to zero.
Calculate Derivative: The derivative of f(x) with respect to x is f′(x). Using the product rule for differentiation, we get: f′(x)=dxd[xln(x)]=ln(x)+x⋅(x1)=ln(x)+1.
Set Derivative Equal to Zero: Now we set the derivative equal to zero to find the critical points: ln(x)+1=0.
Solve for x: To solve for x, we subtract 1 from both sides:ln(x)=−1.
Exponentiate Equation: To get x by itself, we exponentiate both sides of the equation: eln(x)=e−1.
Determine Critical Point Type: Since eln(x) is just x, we have:x=e−1=e1.
Substitute x 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) is increasing for x > 0, and since the critical point x=e1 is the only critical point, this suggests that f(x) has an absolute minimum at x=e1.
Calculate Minimum Value: To find the minimum value of f(x), we substitute x=e1 into the original function:f(e1)=(e1)⋅ln(e1).
Final Answer: We know that ln(e1) is the same as ln(1)−ln(e), which simplifies to 0−1 because ln(1)=0 and ln(e)=1. So, f(e1)=(e1)⋅(−1)=−e1.
Final Answer: We know that ln(e1) is the same as ln(1)−ln(e), which simplifies to 0−1 because ln(1)=0 and ln(e)=1. So, f(e1)=(e1)∗(−1)=−e1. Therefore, the absolute minimum value of the function f(x)=xln(x) is −e1, which corresponds to answer choice (C).
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