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    Hi guys, so I don't get part iv) only. Can someone please help me here? Is there are formula that I need to know for this?
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    (Original post by sienna2266)
    Hi guys, so I don't get part iv) only. Can someone please help me here? Is there are formula that I need to know for this?

    The 'inverse' of f is just a reflection of f in the line y=x. To find this, you swap x and y in its equation. Now the same thing happens to the gradient. You get \dfrac{dy}{dx} turns into \dfrac{dx}{dy} and this is the gradient function of f^{-1}(x)

    So really, \dfrac{d}{dx} f^{-1}(x) = \dfrac{dx}{dy} = \dfrac{1}{ \frac{dy}{dx}} where y=f(x)
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    (Original post by RDKGames)
    The 'inverse' of f is just a reflection of f in the line y=x. To find this, you swap x and y in its equation. Now the same thing happens to the gradient. You get \dfrac{dy}{dx} turns into \dfrac{dx}{dy} and this is the gradient function of f^{-1}(x)

    So really, \dfrac{d}{dx} f^{-1}(x) = \dfrac{dx}{dy} = \dfrac{1}{ \frac{dy}{dx}} where y=f(x)
    Thanks so much!
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    Is my working out correct here?
    So here, at point (1,4) of y=f(x), the gradient is 60 and at point (4,1) of y=f^-1(x), the gradient is 1/60 ?
    Also, is there a quicker/alternative way to do this assuming that I've got the correct working out here?
 
 
 
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