The largest interval, in which the function $f(x) = x^3 + 2x^2 - 1$ is increasing, is:

A function is increasing when the derivative is positive, $f'(x) > 0$.

Given $f(x) = x^3 + 2x^2 - 1$

$f'(x) = 3x^2 + 4x$

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To find critical points, set the derivative equal to zero:

$3x^2 + 4x = 0$

$x(3x + 4) = 0$

$x = 0$ or $x = -\dfrac{4}{3}$

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The critical points divide the number line into three intervals: $(-\infty, -\dfrac{4}{3})$, $(-\dfrac{4}{3}, 0)$, and $(0, \infty)$.

Test $x = -2$ in the interval $(-\infty, -\dfrac{4}{3})$:

$f'(-2) = 3(4) + 4(-2) = 12 - 8 = 4 > 0$

The function is increasing on $(-\infty, -\dfrac{4}{3})$.

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Test $x = -1$ in the interval $(-\dfrac{4}{3}, 0)$:

$f'(-1) = 3(1) + 4(-1) = 3 - 4 = -1 < 0$

The function is decreasing on $(-\dfrac{4}{3}, 0)$.

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Test $x = 1$ in the interval $(0, \infty)$:

$f'(1) = 3(1) + 4(1) = 3 + 4 = 7 > 0$

The function is increasing on $(0, \infty)$.

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The function is increasing on $(-\infty, -\dfrac{4}{3}] \cup [0, \infty)$.