Question

${\boxed{\boxed{\begin{array}{cc}\bf \: if \: \: f(x) = \:\begin{cases} &\sf{1 - |x| } \: \: \: \: \: when \: \: \: |x| \leqslant 0\ \\ \\ &\sf{ |x| - 1 } \: \: \: \: \: when \: \: \: \: |x| > 0 \end{cases} \\ \\ \\ \bf \: and \\ \\ \bf \: g(x) = f(x - 1) + f(x + 1) \\ \\ \\ \sf \: then \: find \: the \: value \: of \: \: \: \displaystyle \: \int_ { - 3}^{5} \rm \: \: g(x) \: dx\end{array}}}}$​

Answer 1

Answer:

Solution:

Given g is the inverse of a function f.

g(x) = f-1x

So fog(x) = x

Differentiate w.r.t.x, we get

f’(g(x)). g’(x) = 1

g’(x) = 1/f’(g(x))

Given f’(x) = 1/(1+x5)

So g’(x) = 1+ g(x)5

= 1+ {g(x)}5

• Hence option (4) is the answer.

Answer 2

Step-by-step explanation:

see the attachment

go through it

it's Isha