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The derivative of a function at a specific point provides information about the function's instantaneous rate of change at that point. For the given problem, the local linearization of the function at the point \(a=3\) is given by \(L(x) = -2x + 5\). This is essentially a tangent line that approximates the function near \(x=3\).

To find \(p(3)\), we evaluate the linearization at \(x=3\):

\[ L(3) = -2(3) + 5 = -6 + 5 = -1 \]

Thus, \(p(3) = -1\). This is because the local linearization should match the actual function value at the point of linearization.

Now, to find \(p'(3)\), we look at the slope of the linearization. The linear function \(L(x) = -2x + 5\) has a slope of -2. Therefore, \(p'(3) = -2\). The slope of the tangent line at \(a=3\) represents the derivative at that point, providing insight into the rate of change of the function at \(x=3\).

Concavity describes the direction in which a function curves. If the second derivative of a function \(p''(x)\) is positive over an interval, the function is concave up on that interval. If \(p''(x)\) is negative, the function is concave down.

In the given exercise, we know that \(p''(3) = 0\) and \(p''(x) < 0\) for \(x < 3\). This indicates that the function is concave down to the left of \(x = 3\). When a function is concave down, it bends downwards like an upside-down bowl. Because of this concavity, the linear approximation you make for \(p(2.79)\) might be too high, or too large.

The point \(x = 3\) where the concavity changes from downwards to upwards is called an inflection point. This means at \(x = 3\), the function changes its curvature which can either be from concave down to concave up or vice versa.

Given \(p''(x) > 0\) for \(x > 3\), the function is concave up to the right of \(x = 3\). This setup allows us to visually predict how the function behaves—concave down to the left and up to the right of the inflection point.

Linear approximation involves using the tangent line at a specific point to estimate values of the function near that point. This technique is valuable when the exact value of the function is complex to compute.

In the exercise, we use the linear approximation formula which is:

\[ L(x) = p'(a)(x - a) + p(a) \]

Here, \(a = 3\), \(p'(3) = -2\), and \(p(3) = -1\). To estimate \(p(2.79)\), we substitute these values:

\[ L(2.79) = -2 * (2.79 - 3) + (-1) \]

Simplifying, we get:

\[ L(2.79) = -2 * (-0.21) - 1 = 0.42 - 1 = -0.58 \]

Thus, according to the linear approximation, \(p(2.79)\) is approximately -0.58.

This linear approximation is particularly helpful when working near the point where you have the linear representation. However, remember this method might not be perfectly accurate due to the function's concavity. In this case, since the function is concave down to the left of \(x = 3\), we know our approximation might be slightly too high.