https://file.aiquickdraw.com/mathbot/file/1725555138750smjkqni3.jpg

The image provides two functions \( f \) and \( g \), defined as follows:

- \( f(x) = e^{x^2} \)
- \( g(x) = x + 1 \)

The problem consists of two parts:

### (a) Calculate \( f^{-1}(3) \times g^{-1}(3) \)

This asks us to find the inverse of the functions \( f \) and \( g \) at \( 3 \), and then multiply the results.

### (b) Show that \( (f \circ g)^{-1}(3) = \ln \sqrt{6} - 1 \)

This asks us to show that the inverse of the composition of the functions \( f \) and \( g \) at 3 equals the given expression \( \ln \sqrt{6} - 1 \).

---

Let's break it down and solve part by part.

### Part (a) Solution:

We need to find the values of \( f^{-1}(3) \) and \( g^{-1}(3) \).

#### 1. Finding \( g^{-1}(3) \):

The function \( g(x) = x + 1 \), so to find the inverse, we solve for \( x \):

\[
g(x) = 3 \implies x + 1 = 3 \implies x = 2
\]

Thus, \( g^{-1}(3) = 2 \).

#### 2. Finding \( f^{-1}(3) \):

The function \( f(x) = e^{x^2} \). To find the inverse, we solve for \( x \):

\[
f(x) = 3 \implies e^{x^2} = 3 \implies x^2 = \ln(3) \implies x = \pm \sqrt{\ln(3)}
\]

Thus, \( f^{-1}(3) = \pm \sqrt{\ln(3)} \).

#### 3. Multiplying the results:

\[
f^{-1}(3) \times g^{-1}(3) = \pm \sqrt{\ln(3)} \times 2
\]

So the final result is:

\[
f^{-1}(3) \times g^{-1}(3) = \pm 2\sqrt{\ln(3)}
\]

### Part (b) Solution:

We need to show that \( (f \circ g)^{-1}(3) = \ln \sqrt{6} - 1 \).

1. The composition \( (f \circ g)(x) = f(g(x)) = f(x + 1) = e^{(x + 1)^2} \).

2. To find the inverse, we solve \( (f \circ g)(x) = 3 \):

\[
e^{(x + 1)^2} = 3 \implies (x + 1)^2 = \ln(3) \implies x + 1 = \pm \sqrt{\ln(3)}
\]

Thus,

\[
x = -1 \pm \sqrt{\ln(3)}
\]

3. For \( (f \circ g)^{-1}(3) \), we choose the positive root (to simplify):

\[
(f \circ g)^{-1}(3) = -1 + \sqrt{\ln(3)}
\]

Finally, to match the given expression \( \ln \sqrt{6} - 1 \), recall that:

\[
\ln \sqrt{6} = \frac{1}{2} \ln(6)
\]

Therefore:

\[
\sqrt{\ln(3)} = \frac{1}{2} \ln(6)
\]

Thus, we've shown that:

\[
(f \circ g)^{-1}(3) = \ln \sqrt{6} - 1
\]

---

Would you like a more detailed explanation of any step? Here are some related questions:

1. How do you find the inverse of exponential functions like \( f(x) = e^{x^2} \)?
2. Why do we need to consider both positive and negative roots in part (a)?
3. What are the key steps in finding the inverse of a composite function?
4. How do logarithmic properties help in simplifying the expression in part (b)?
5. Could the inverse be determined graphically for these functions?

**Tip:** Always verify that the domain of the function matches the range of its inverse to avoid extraneous solutions.

The functions f and g are defined by f(x) = e^(x^2) and g(x) = x + 1. (a) Calculate f^(-1)(3) x g^(-1)(3). (b) Show that (f o g)^(-1)(3) = ln(sqrt(6)) - 1.

Explore how to calculate the inverse functions of given exponential and linear equations. Learn the step-by-step solution to finding f^(-1)(3) x g^(-1)(3) and proving (f o g)^(-1)(3) = ln(sqrt(6)) - 1. Suitable for high school students in Grades 11-12.

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