#### To determine

**To find:**

(a) To show that g''x= -f''gxf'gx3

(b) g=f-1 is concave downward.

#### Answer

(a) Proved.

(b) g=f-1 is concave downward.

#### Explanation

Given f is one to one and f-1=g.

By Theorem 7,f-1'x=1f'f-1x=1f'gx ……………….. (A)

Now,g''x=f-1''x=-f'gx'f'gx2 (by the division rule)

=-f''gxg'xf'gx2 (by the chain rule)

=-f''gx(f-1)'xf'gx2 since g=f-1

= -f''gxf'gx2f'gx (by equation (A))

= -f''gxf'gx3 .

Hence proved.

b)

We know that f is increasing and concave upward.

Therefore,f't, f''t>0, ∀ t Domain. If gx Domain, then f'gx>0 and f''gx>0. Then by using the above formula, g''x<0 g=f-1 is concave downward.

**Note**: A sufficient condition for a function g to be concave downwad is that:

If g''>0, then g is concave downward.

**Conclusion:**

(a) Proved.

(b) g=f-1 is concave downward.