The interval in which the function y = x³ + 5x² – 1 is decreasing, is

The correct option is (−10/3, 0).

To find the interval in which the given function is decreasing, we analyze the sign of its first derivative. A function is strictly decreasing on an interval where its first derivative with respect to $x$ is negative.

The given function is:

$$y=x^3+5x^2-1$$

First, we find the first derivative of the function,
$$\frac{dy}{dx}$$
using the power rule of differentiation:

$$\frac{dy}{dx}=3x^2+10x$$

For the function to be decreasing, we must have:
$$\frac{dy}{dx}<0$$

Substituting the derivative, we get the quadratic inequality:

$$3x^2+10x<0$$

We can factor out $x$ from the expression:

$$x(3x+10)<0$$

To solve this inequality, we identify the critical points where the expression equals zero:
$$x=0$$
and
$$3x+10=0\Rightarrow x=-\frac{10}{3}$$

These critical points divide the real number line into three intervals:
$(-\infty ,-\frac{10}{3})$,
$(-\frac{10}{3},0)$, and
$(0,\infty )$.

We test the sign of the product $x(3x+10)$ in each interval:

1. For $x<-\frac{10}{3}$, both factors $x$ and $3x+10$ are negative, so their product is positive ($>0$).

2. For $-\frac{10}{3}<x<0$, the factor $x$ is negative and $3x+10$ is positive, so their product is negative ($<0$).

3. For $x>0$, both factors are positive, so their product is positive ($>0$).

Thus, the inequality $x(3x+10)<0$ is satisfied when $x$ lies in the interval:

$$(-\frac{10}{3},0)$$

Therefore, the function is decreasing in the interval (−10/3, 0).