It is given that $\mathrm{f}(x) = 2x^6 - 4x^4 - 6x^2 - 7$. The diagram shows the curve with equation $y = \mathrm{f}(x)$ for $x \ge 0$. The curve crosses the positive $x$-axis at $x = \alpha$.

1. Set the function to zero to find the positive $x$-intercept.

At the $x$-axis, $\mathrm{f}(x) = 0$.
$2x^6 - 4x^4 - 6x^2 - 7 = 0$

2. Solve the equation for $\alpha$ using a graphing calculator.

Since $\alpha$ is the positive root, we use a graphing calculator to solve for $x > 0$:
$x \approx 1.8036$
Therefore, $\alpha = 1.804$ (correct to 3 decimal places).

1. Show that $\mathrm{f}(x) = \mathrm{f}(-x)$.

$\begin{aligned} \mathrm{f}(-x) &= 2(-x)^6 - 4(-x)^4 - 6(-x)^2 - 7 \\ &= 2x^6 - 4x^4 - 6x^2 - 7 \\ &= \mathrm{f}(x) \end{aligned}$

2. Use the symmetry of the graph to deduce the number of real roots.

Since $\mathrm{f}(x) = \mathrm{f}(-x)$, the graph of $y = \mathrm{f}(x)$ is an even function and is symmetrical about the $y$-axis. This implies that if $x$ is a root, $-x$ is also a root.
From the given graph, there is exactly one root for $x \ge 0$, which is $x = \alpha$.
Due to symmetry, there is exactly one root for $x < 0$, which is $x = -\alpha$.
Thus, the equation $\mathrm{f}(x) = 0$ has exactly two real roots, $x = \alpha$ and $x = -\alpha$.

3. Deduce the nature of the remaining roots.

Since $\mathrm{f}(x)$ is a polynomial of degree 6, the equation $\mathrm{f}(x) = 0$ must have exactly 6 roots in total.
Because only 2 of these roots are real, the remaining 4 roots must be non-real complex numbers.
(Note: As the coefficients are real and the function is even, these 4 complex roots occur in conjugate pairs and in $\pm$ pairs).

1. Establish the condition for the stationary points of $y = \mathrm{g}(x)$

Stationary points occur when $\mathrm{g}'(x) = 0$.
Given $\mathrm{g}'(x) = \mathrm{f}(x)$, this implies we need to solve $\mathrm{f}(x) = 0$.

2. Find the $x$-coordinates of the stationary points using earlier results

From (i), solving $2x^6 - 4x^4 - 6x^2 - 7 = 0$ using a graphing calculator yields the positive root $x = \alpha \approx 1.804$.
From (ii), $\mathrm{f}(x)$ is an even function ($\mathrm{f}(x) = \mathrm{f}(-x)$), so there is a corresponding real negative root at $x = -\alpha \approx -1.804$.
Thus, the $x$-coordinates of the stationary points are $x \approx 1.804$ and $x \approx -1.804$.

3. Determine the nature of the stationary point at $x = \alpha$

From the given graph for $x \ge 0$, as $x$ increases through $\alpha$, the value of $\mathrm{f}(x)$ changes from negative to positive.
Since $\mathrm{g}'(x) = \mathrm{f}(x)$, the gradient of $\mathrm{g}(x)$ changes from negative to positive ($- \to 0 \to +$).
By the first derivative test, $y = \mathrm{g}(x)$ has a local minimum at $x = 1.804$.

4. Determine the nature of the stationary point at $x = -\alpha$

By the symmetry of the graph (reflected across the $y$-axis since it is an even function), as $x$ increases through $-\alpha$, the value of $\mathrm{f}(x)$ changes from positive to negative.
Thus, the gradient of $\mathrm{g}(x)$ changes from positive to negative ($+ \to 0 \to -$).
By the first derivative test, $y = \mathrm{g}(x)$ has a local maximum at $x = -1.804$.

1. State the condition for the graph of $y = \mathrm{g}(x)$ to be concave upwards.

For the graph of $y = \mathrm{g}(x)$ to be concave upwards, we require $\mathrm{g}''(x) > 0$.

Given $\mathrm{g}'(x) = \mathrm{f}(x)$ for all real $x$, differentiating both sides gives $\mathrm{g}''(x) = \mathrm{f}'(x)$.

Thus, we need to find the range of $x$ for which $\mathrm{f}'(x) > 0$.

2. Find $\mathrm{f}'(x)$ and formulate the inequality.

Given $\mathrm{f}(x) = 2x^6 - 4x^4 - 6x^2 - 7$,
$\mathrm{f}'(x) = 12x^5 - 16x^3 - 12x$

Set the derivative strictly greater than zero:
$12x^5 - 16x^3 - 12x > 0$
$4x(3x^4 - 4x^2 - 3) > 0$

3. Solve the inequality to find the required values of $x$.

Find the roots of $3x^4 - 4x^2 - 3 = 0$ by letting $u = x^2$:
$u = \frac{4 \pm \sqrt{(-4)^2 - 4(3)(-3)}}{2(3)} = \frac{4 \pm \sqrt{52}}{6} = \frac{2 \pm \sqrt{13}}{3}$
Since $x^2 \ge 0$ for all real $x$, we reject the negative value.
Thus, $x^2 = \frac{2 + \sqrt{13}}{3} \approx 1.8685$.
This gives $x \approx \pm 1.3669$.
The real roots of $\mathrm{f}'(x) = 0$ are $x = -1.3669$, $x = 0$, and $x = 1.3669$.

Testing intervals around these roots (or by sketching the curve/using a graphing calculator) shows that $4x(3x^4 - 4x^2 - 3) > 0$ in the intervals $(-1.3669, 0)$ and $(1.3669, \infty)$.

Therefore, the graph is concave upwards for:
$-1.37 < x < 0 \quad \text{or} \quad x > 1.37 \text{ (to 3 s.f.)}$