Math  /  Calculus

QuestionConsider the function f(x)=x5ln(x),15x8f(x)=x-5 \ln (x), \quad \frac{1}{5} \leq x \leq 8. The absolute maximum value is \square and this occurs at xx equals \square The absolute minimum value is \square and this occurs at xx equals \square

Studdy Solution
Compare the values to determine the absolute maximum and minimum.
Evaluate:
1. f(15)=155ln(15) f\left(\frac{1}{5}\right) = \frac{1}{5} - 5 \ln\left(\frac{1}{5}\right)
2. f(5)=55ln(5) f(5) = 5 - 5 \ln(5)
3. f(8)=85ln(8) f(8) = 8 - 5 \ln(8)

Calculate these values to compare:
1. f(15)=15+5ln(5) f\left(\frac{1}{5}\right) = \frac{1}{5} + 5 \ln(5) (since ln(15)=ln(5)\ln\left(\frac{1}{5}\right) = -\ln(5))
2. f(5)=55ln(5) f(5) = 5 - 5 \ln(5)
3. f(8)=85ln(8) f(8) = 8 - 5 \ln(8)

Determine which is the largest and which is the smallest.
The absolute maximum value is f(8)=85ln(8) f(8) = 8 - 5 \ln(8) and this occurs at x=8 x = 8 .
The absolute minimum value is f(5)=55ln(5) f(5) = 5 - 5 \ln(5) and this occurs at x=5 x = 5 .

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