#### To determine

**To find:**

The value of (f−1)′(1) for the given expression.

**Solution:** 12.

#### Explanation

**Given:**

f(x)=x+x2+ex

**Formulae used:**

(f−1)′=1f′(f−1(x))

Consider f(x)=x+x2+ex

The domain of f−1(x) is the range of f(x) and the range of f−1(x) is the domain of f(x)

The domain of f−1(x) is 1. It means that the range of the f(x) is 1

Therefore,

f(x)=x+x2+exx+x2+ex=1x+x2+ex=1

Only x=0 is satisfied the above equation.

Thus, the domain of f(x) is 0. It means that the range of f−1(x) is also 0

Hence, use the formulae of (f−1)′=1f′(f−1(x))

Now according to the given expression, the value of the expression (f−1)′(1)

1f′(f−1(π4))=1f′(0)f′(x)=1+2x+exf′(0)=21f′(f−1(π4))=12

**Conclusion:** The value of expression (f−1)′(π4) is 12.