f ( x ) = x ∗ e x f(x)=x*e^x f ( x ) = x ∗ e x
f ′ ( x ) = e x + x ∗ e x f'(x)=e^x+x*e^x f ′ ( x ) = e x + x ∗ e x
f − 1 ( x ) − inverse function f ( x ) f^{-1}(x) -\text{inverse function }f(x) f − 1 ( x ) − inverse function f ( x )
( f − 1 ( x ) ) ′ = 1 f ′ ( f − 1 ( x ) ) (f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))} ( f − 1 ( x ) ) ′ = f ′ ( f − 1 ( x )) 1
f − 1 ( x ) : x = y ∗ e y f^{-1}(x):x=y*e^y f − 1 ( x ) : x = y ∗ e y
x = y ∗ e y − this function does not allow you to explicitly express y in x x=y*e^y - \text{this function does not allow you to explicitly express y in x} x = y ∗ e y − this function does not allow you to explicitly express y in x
x ≈ y ∗ ( 1 + y ) x\approx{y*(1+y) } x ≈ y ∗ ( 1 + y )
y ≈ − 1 + 1 + 4 ∗ x 2 y\approx\frac{-1+\sqrt{1+4*x}}{2} y ≈ 2 − 1 + 1 + 4 ∗ x
f − 1 ( 0.5 ) ≈ 0.37 f^{-1}(0.5)\approx0.37 f − 1 ( 0.5 ) ≈ 0.37
( f − 1 ( 0.5 ) ) ′ = 1 f ′ ( 0.37 ) = 1 e 0.37 + 0.37 ∗ e 0.37 ≈ 0.504 (f^{-1}(0.5))'=\frac{1}{f'(0.37)}=\frac{1}{e^{0.37}+0.37*e^{0.37}}\approx 0.504 ( f − 1 ( 0.5 ) ) ′ = f ′ ( 0.37 ) 1 = e 0.37 + 0.37 ∗ e 0.37 1 ≈ 0.504
Answer: 0.504
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