The diagram below shows the curve $y = \text{f}'(x)$. It has turning points at $(0, 6)$, $(5, 0)$ and $(7, 7)$ and intersects the $x$-axis at $(-3, 0)$ and $(8, 0)$.

1. State the $x$-coordinates of the stationary points by finding where $\text{f}'(x) = 0$.

Stationary points occur when $\text{f}'(x) = 0$, which corresponds to the $x$-intercepts on the $y = \text{f}'(x)$ graph.
From the given graph, the $x$-intercepts are $x = -3$, $x = 5$, and $x = 8$.

2. Determine the nature of the stationary point at $x = -3$.

Just before $x = -3$, the curve $y = \text{f}'(x)$ is below the $x$-axis ($\text{f}'(x) < 0$).
Just after $x = -3$, the curve is above the $x$-axis ($\text{f}'(x) > 0$).
Since the gradient changes from negative to positive, $x = -3$ is a local minimum point.

3. Determine the nature of the stationary points at $x = 5$ and $x = 8$.

At $x = 5$, the curve $y = \text{f}'(x)$ touches the $x$-axis and remains above it. Since $\text{f}'(x) > 0$ on both sides, the gradient does not change sign, so $x = 5$ is a stationary point of inflexion.

At $x = 8$, the curve crosses from above the $x$-axis ($\text{f}'(x) > 0$) to below the $x$-axis ($\text{f}'(x) < 0$). Since the gradient changes from positive to negative, $x = 8$ is a local maximum point.

1. Relate the concavity of $y = \text{f}(x)$ to the gradient of $y = \text{f}'(x)$

For the curve $y = \text{f}(x)$ to be concave downwards, we require $\text{f}''(x) < 0$.
This means the gradient of the curve $y = \text{f}'(x)$ must be negative, which occurs when the graph of $y = \text{f}'(x)$ is strictly decreasing.

2. Identify the relevant intervals from the graph

From the given graph, $y = \text{f}'(x)$ is decreasing between the turning points $(0, 6)$ and $(5, 0)$, and after the turning point $(7, 7)$.
Therefore, the required range of values is $0 < x < 5$ and $x > 7$.

1. Identify the sign of $\text{f}'(x)$ from the graph and deduce the behavior of $\text{f}(x)$ for $-3 < x < 5$.

From the given graph, the curve $y = \text{f}'(x)$ lies above the $x$-axis for $-3 < x < 5$. This means $\text{f}'(x) > 0$ in this interval, indicating that the function $\text{f}(x)$ is strictly increasing.

2. Evaluate the student's claim based on the initial value at $x = -3$.

Given that the curve passes through $(-3, 7)$, we have $\text{f}(-3) = 7$.
Since $\text{f}(x)$ is strictly increasing for $-3 < x < 5$, it follows that $\text{f}(x) > \text{f}(-3)$ for all $x$ in this interval.
Therefore, $\text{f}(x) > 7$ for $-3 < x < 5$, and the student is right.