Question #162391

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


1
Expert's answer
2021-02-24T07:09:21-0500

f(x)=xexf(x)=x*e^x

f(x)=ex+xexf'(x)=e^x+x*e^x

f1(x)inverse function f(x)f^{-1}(x) -\text{inverse function }f(x)

(f1(x))=1f(f1(x))(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}

f1(x):x=yeyf^{-1}(x):x=y*e^y

x=yeythis 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}

xy(1+y)x\approx{y*(1+y) }

y1+1+4x2y\approx\frac{-1+\sqrt{1+4*x}}{2}

f1(0.5)0.37f^{-1}(0.5)\approx0.37

(f1(0.5))=1f(0.37)=1e0.37+0.37e0.370.504(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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