Integration maths help pleasee

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?

Reply 1

RDKGames

Study Forum Helper

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)

Reply 2

sienna2266

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! :smile:

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?

(edited 6 years ago)

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Integration maths help pleasee

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