Practicing Success
Practicing Success
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The minimum value of $x^x (x > 0)$ is |
at x = 1 at x = e at $x = e^{–1}$ none of these |
at $x = e^{–1}$ |
Let $y = x^x \log y = x \log x$ $(\log y) = 1 + \log x$ and $(\log y) = x^{–1}$ minimum value of y or $\log y$ $(\log y) = 0\, 1 + \log x = 0$ $x = e^{–1}$ again $x = e^{–1}$ $(\log y ) = e > 0$ minimum value of at $x = e^{–1}$ |
