Q.Find the intervals in which the function $f$ given by $f(x) = x^3 + \frac{1}{x^3}, x \neq 0$ is

We are asked about the nature of $f(x) = x^3 - 3x^2 + 4x$ on the real numbers — specifically, whether it is strictly decreasing, decreasing, increasing, or constant. The key is to examine the sign of the derivative. A function is increasing on an interval if its derivative is non-negative there, and strictly increasing if the derivative is positive (except possibly at isolated points). Similarly for decreasing.

1. Compute the derivative.

This is a quadratic. Its sign tells us the slope of $f$ at every point.

2. Determine the sign of $f'(x)$.

Check the discriminant:

Since $\Delta < 0$ and the leading coefficient $3 > 0$, the quadratic is always positive for all real $x$.

Thus $f'(x) > 0$ for every $x \in \mathbb{R}$.

3. Interpret the result.

If the derivative is positive everywhere, the function is strictly increasing on the entire real line. It never levels off (no flat spots) and never decreases. …

Step 1 — Fix the domain first.

$\log(1+x)$ is defined only for $1+x>0$, i.e. $x > -1$. Any answer containing $x \le -1$ (options A, C, D) is therefore impossible before we even differentiate — but let us prove (B) properly.

Step 2 — Differentiate.

For the second term use the quotient rule:

Hence

Step 3 — Combine into one fraction.

Step 4 — Read the sign.

• $x^{2} \ge 0$ always. …

We are asked about the behavior of $f(x)=x^2-x+1$ on the open interval $(0,1)$. The key is to determine whether it is always increasing, always decreasing, or changes direction. Since this is a quadratic, its graph is a parabola — and the sign of its derivative tells us where it rises and falls.

Why this approach works:

For a differentiable function, monotonicity (increasing or decreasing) on an interval is determined by the sign of its derivative. If the derivative is positive everywhere, the function is strictly increasing; if negative everywhere, strictly decreasing. If the derivative changes sign, the function is neither.

1. Find the derivative.

This is a linear function, so it can change sign at most once.

2. Find where the derivative is zero. Set $f'(x)=0$:

This critical point lies inside $(0,1)$.

3. Test the sign of $f'(x)$ on either side of $x=\frac12$.

   • For $x < \frac12$, say $x=0$: $f'(0) = -1 < 0$ → function is decreasing.
   • For $x > \frac12$, say $x=0.75$: $f'(0.75) = 1.5 - 1 = 0.5 > 0$ → function is increasing.

4. Interpret the result on $(0,1)$. …

We want to determine when

is decreasing for all $x$.

A function is decreasing on an interval if its derivative is negative everywhere on that interval.

Here, the domain excludes points where the denominator is zero, but the condition “decreasing for all $x$” means for every $x$ where $f$ is defined, $f'(x) < 0$.

The key is to compute $f'(x)$ and see what inequality on $a,b,c,d$ makes it always negative.

1. Differentiate using the quotient rule Let $u = a\sin x + b\cos x$ and $v = c\sin x + d\cos x$. Then

The derivative is

2. Compute the numerator

Subtract:

3. Expand both products First product:

Second product:

Notice $\cos x\sin x = \sin x\cos x$.

4. Subtract term by term

   • The $ac$ terms: $ac\cos x\sin x - ac\sin x\cos x = 0$.
   • The $bd$ terms: $-bd\sin x\cos x - (-bd\cos x\sin x) = -bd\sin x\cos x + bd\sin x\cos x = 0$.
   • The $ad$ terms: $ad\cos^2 x - (-ad\sin^2 x) = ad(\cos^2 x + \sin^2 x) = ad$.
   • The $bc$ terms: $-bc\sin^2 x - bc\cos^2 x = -bc(\sin^2 x + \cos^2 x) = -bc$.

   So the numerator simplifies beautifully to

   $$$$