A calculus problem by A Former Brilliant Member
f ( x ) = − 4 e 1 − x + 1 + x + 2 x 2 + 3 x 3
If g ( x ) is the inverse of f ( x ) above, then find g ′ ( 6 − 7 ) 2 , where g ′ is the first derivative of g .
The answer is 10.
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2 solutions
Its very obvious to use the relation g(f(x)) = x where g is the inverse of x
It can be seen by differentiating f(x) its an Injective as well as Surjective function. Hence inverse of function exists
Differentiating the equation wrt g'(f(x)) = 1/f'(x)
By observation f(1) = -7/6
Put x = 1 g'(-7/6) = 1/f'(1)
Simply differentiate f(x) and put x = 1 to get the required value
exact bro !!
yaaer ek Q poochne tha electrostaics ka ........ A square of side a is taken a charge +q is placed directly above the center at a distance a/2 of the square and charge -q at a distance a directly below the corner of sqruare , what is the net flux through the square... @Prakhar Bindal
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Bhai q dhang se likha nhi hai
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a dekhiyo bhai !
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@A Former Brilliant Member – Bhai jo charge +q hai wo kahan par hai square ke andar ya bahar?
-q kahan par hai
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@Prakhar Bindal – no thing other than this mentioned m'bro . i had confusions in the scenario too ! bhai isme hi batana tha answer @Prakhar Bindal
If g ( x ) is the inverse of f ( x ) , then the composition g ( f ( x ) ) = x holds. Differentiating both sides with respect to x yields f ′ ( x ) ∗ g ′ ( f ( x ) ) = 1 , or g ′ ( f ( x ) ) = 1 / f ′ ( x ) . . . ( i ) . Next, we need to find the value of x such that f ( x ) = − 7 / 6 , which upon observation using x = 1 :
f ( 1 ) = − 4 e x p [ ( 1 − 1 ) / 2 ] + 1 + 1 + 1 / 2 + 1 / 3 = − 2 + 5 / 6 = − 7 / 6 .
which differentiating f(x) produces:
2 e x p [ ( 1 − x ) / 2 ] + 1 + x + x 2 ;
and g ′ ( − 7 / 6 ) = 1 / f ′ ( 1 ) = 1 / ( 2 + 1 + 1 + 1 ) = 1 / 5 ;
thus, the quantity 2 / g ′ ( − 7 / 6 ) = 2 / ( 1 / 5 ) = 1 0 .