3. (a) At the front of a building there are five garage doors - HSC - SSCE Mathematics Extension 1 - Question 3 - 2010 - Paper 1

If we consider the two red doors as a single unit, we can treat them as one 'block'. Therefore, we have four units to arrange: the 'red block', green, blue, and orange.

The arrangements are:

$4! = 24$

Within the 'red block', the two red doors can be arranged among themselves in:

$2! = 2$

Therefore, the total arrangements are:

$4! \times 2! = 24 \times 2 = 48$

So, there are 48 arrangements where the two red doors are next to each other.

To find the points of inflexion, we need to determine where the second derivative of the function changes sign. We begin by finding the first derivative:

$f'(x) = 2xe^{x^2}$

Now, finding the second derivative using the product rule:

$f''(x) = 2e^{x^2} + 2x(2xe^{x^2}) = 2e^{x^2}(1 + 2x^2)$

Setting the second derivative equal to zero:

$2e^{x^2}(1 + 2x^2) = 0$

Since $e^{x^2} > 0$, we can solve:

For a function to possess an inverse, it must be one-to-one (bijective). The function $f(x) = e^{x^2}$ is not one-to-one for its entire domain since it is symmetric about the y-axis. This means that for any positive $y$, there are two corresponding $x$ values: one positive and one negative. Therefore, we restrict the domain, typically to $x \geq 0$, to ensure each output corresponds to a unique input.

Given that the original function is $y = e^{x^2}$, we can express it as:

The domain of $f^{-1}(x)$ corresponds to the range of $f(x)$ restricted to $x \geq 0$. Since $f(x)$ can take values from 1 to infinity when $x \geq 0$, the domain of the inverse function is:

$\text{Domain of } f^{-1}(x): [1, \infty)$.

To sketch the curve $y = f^{-1}(x)$, we recognize that it represents the function $y = \sqrt{\ln(x)}$. The curve starts from the point (1,0) when $x=1$ and increases gradually as $x$ increases, with the value of $y$ increasing without bound as $x$ approaches infinity. The graph will be located in the first quadrant, curving upwards.

To prove there is a solution, we evaluate the function $g(x) = x - e^{-x^2}$. We compute:

g(0.6) = 0.6 - e^{-0.36} \approx 0.6 - 0.6977 \approx -0.0977\ $g(0.7) = 0.7 - e^{-0.49} \approx 0.7 - 0.6122 \approx 0.0878$.

Since $g(0.6) < 0$ and $g(0.7) > 0$, by the Intermediate Value Theorem, there exists at least one solution in the interval [0.6, 0.7].