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.