Question

If f : R → R, g : R → R defined by f(x) = 3x - 2, g(x) = x² + 1, then findi. (gof⁻¹) (2)ii. (gof) (x - 1)

Answer 1

Given,

f : R → R

f(x) = 3x - 2

g : R → R

g(x ) = x² + 1

Now,

(i) (gof⁻¹)

First find gof

So,

(gof)(x) = g(f(x))

= g ( 3x - 2 )

= (3x - 2 )² + 1

= 9x² + 4 - 12x + 1

= 9x² - 12x + 5

Let gof = y

y = (3x - 2 )² + 1

y + 1 =( 3x - 2)²

√(y + 1 ) = 3x - 2

√(y+1) + 2 / 3 = x.

Now,
gof'(y) = $\frac{\sqrt{(y+1)} + 2}{3}$

So,

gof'(x) = $\frac{\sqrt{(x+1)} + 2}{3}$

(ii) gof(x - 1 )

= [ 3(x-1) - 2 ]² + 1

= [ 3x - 3 - 2 ]² + 1

= [ 3x - 5 ] ² + 1

= 9x² + 25 - 30x + 1

= 9x² + 26 - 30x.

Hope helped! ^^