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Sagot :
Given:
[tex]f(0)=4_{}[/tex][tex]f^{\prime}(0)=\frac{1}{3}[/tex]We get the point (0,4) from f(0)=4.
we get the slope m=1/3 from f'(0)=1/3.
The point-slope formula is
[tex]y-y_1=m(x-x_1)[/tex][tex]\text{ Substitute }y_1=4,x_1=0\text{ and m=}\frac{1}{3},\text{ we get}[/tex][tex]y-4=\frac{1}{3}(x-0)[/tex][tex]y-4=\frac{1}{3}x[/tex][tex]\text{Let y=f(x) and substitute x=f}^{-1}(y)[/tex][tex]y-4=\frac{1}{3}f^{-1}(y)[/tex][tex]3(y-4)=f^{-1}(y)[/tex]Replace y by x, we get
[tex]f^{-1}(x)=3(x-4)[/tex][tex]Let\text{ }f^{-1}(x)=h(x)[/tex][tex]h(x)=3(x-4)[/tex]Differentiate with respect to x, we get
[tex]h^{\prime}(x)=3[/tex]Hence the answer is
[tex]h^{\prime}(4)=3[/tex]
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