Example 5. Investigate the monotonicity and extrema of the function $y=2 x^{3}+3 x^{2}-12 x+1$. Also, find the maximum and minimum values of this function on the interval $[-1 ; 5]$.
Solution. We find \( y' \) and investigate its sign:

\[
y' = 6x^2 + 6x - 12 = 6(x-1)(x+2)
\]

The signs of \( y' \) and the intervals of monotonicity are shown in the diagram (Fig. 7.6); \( x = -2 \) and \( x = 1 \) are stationary points, \( x = 1 \) is a point of minimum, and \( x = -2 \) is a point of maximum. Here, \( y_{\min} = y(1) = -6 \), \( y_{\max} = y(-2) = 21 \). These values are extreme. We also compute \( y(-1) = 14 \) and \( y(5) = 266 \).

Answer. The function is monotonically increasing in \( (-\infty, -2) \cup (1, +\infty) \), decreasing in \( (-2, 1) \), has a maximum at \( x = -2 \), \( y_{\max} = 21 \); has a minimum at \( x = 1 \), \( y_{\min} = -6 \) - the smallest value; the largest value is \( y(5) = 266 \).

## 9.3. Resolving Indeterminate Forms

\( 1^{\circ} \). Indeterminate forms of the types \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \)

Let the functions \( f(x) \) and \( g(x) \) be continuous and differentiable in some neighborhood of the point \( x = a \), except possibly at the point \( a \) itself (where \( a \) can be a finite number or \( +\infty, -\infty, \infty \)).

Assume that

\[
\text{1) } \lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0 \quad \text{or} \quad \text{2) } \lim_{x \to a} f(x) = \lim_{x \to a} g(x) = \infty
\]

L'Hôpital's Rule. If the limit \( \lim_{x \to a} \frac{f'(x)}{g'(x)} = A \) exists, then the limit \( \lim_{x \to a} \frac{f(x)}{g(x)} = A \) also exists.

L'Hôpital's Rule allows the use of derivatives to resolve indeterminate forms: in case 1) of the form \( \frac{0}{0} \), and in case 2) of the form \( \frac{\infty}{\infty} \).

If necessary, L'Hôpital's Rule can be applied repeatedly or several times under the appropriate conditions on \( f'(x) \) and \( g'(x) \), \( f''(x) \) and \( g''(x) \), etc.

For example, let's compute the limits: 1) \( \lim_{x \to 0} \frac{e^x - 1}{x} \); 2) \( \lim_{x \to +\infty} \frac{e^x - 1}{x^2} \).

In the first case, we have an indeterminate form of the type \( \frac{0}{0} \), and in the second case, of the type \( \frac{\infty}{\infty} \). In both cases, the conditions for applying L'Hôpital's Rule are satisfied. We will denote the transition to derivatives in the course of solving the examples with the symbol "!" and proceed to the calculations, noting in advance that in the second example, L'Hôpital's Rule can be applied twice.

1) \( \lim_{x \to 0} \frac{e^x - 1}{x} = \left( \frac{0}{0} \right) \stackrel{!}{=} \lim_{x \to 0} \frac{(e^x - 1)'}{x'} = \lim_{x \to 0} \frac{e^x}{1} = 1 \).
2) \( \lim_{x \to +\infty} \frac{e^x - 1}{x^2} = \left( \frac{\infty}{\infty} \right) \stackrel{!}{=} \lim_{x \to +\infty} \frac{e^x}{2x} = \left( \frac{\infty}{\infty} \right) \stackrel{!}{=} \lim_{x \to +\infty} \frac{e^x}{2} = +\infty \).

\( 2^{\circ} \). Using L'Hôpital's Rule, we can resolve other indeterminate forms: \( \infty, -\infty, 0 \cdot \infty, \infty^0, 1^\infty, 0^0 \). For this, the initial expressions must be brought to the form \( \frac{f(x)}{g(x)} \).

In the following examples, the necessary transformations will be performed during the solution. Note that sometimes it is necessary to get rid of "hindering" factors - this is done using the theorem on the limit of a product.

## Examples with Solutions