Practicing Success
Practicing Success
CUET Preparation Today
The maximum value of x–x is |
ee–1 \(e^\frac{1}{e}\) \(e^\frac{-1}{e}\) \(e^\frac{-4}{e}\) |
\(e^\frac{1}{e}\) |
f′(x) = x−x(−logx−1) = −x−x(logx+1) f′(x) = 0 when x = $\frac{1}{e}$ f′(x) > 0 when 0 < x < $\frac{1}{e}$ f′(x) < 0 when x > $\frac{1}{e}$ f(x) is maximum at x = $\frac{1}{e}$ $f(\frac{1}{e})=(e)^{\frac{1}{e}}$ |
