Question #162391

The derivate of inv function of f:|0,1 |➡️R f(x) =xe^x at x=0.5 is


Expert's answer

f(x)=x∗exf(x)=x*e^x

f′(x)=ex+x∗exf'(x)=e^x+x*e^x

f−1(x)−inverse function f(x)f^{-1}(x) -\text{inverse function }f(x)

(f−1(x))′=1f′(f−1(x))(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}

f−1(x):x=y∗eyf^{-1}(x):x=y*e^y

x=y∗ey−this function does not allow you to explicitly express y in xx=y*e^y - \text{this function does not allow you to explicitly express y in x}

x≈y∗(1+y)x\approx{y*(1+y) }

y≈−1+1+4∗x2y\approx\frac{-1+\sqrt{1+4*x}}{2}

f−1(0.5)≈0.37f^{-1}(0.5)\approx0.37

(f−1(0.5))′=1f′(0.37)=1e0.37+0.37∗e0.37≈0.504(f^{-1}(0.5))'=\frac{1}{f'(0.37)}=\frac{1}{e^{0.37}+0.37*e^{0.37}}\approx 0.504

Answer:0.504





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