Question

Let $\phi(x) = (f(x))^3-3(f(x))^2 + 4f(x) + 5x + 3 \sin x + 4 \cos x \forall x \in R$, where $f(x)$ is a differentiable function $\forall x \in R$, then

• A. $\phi$ is increasing whenever f is increasing
• B. $\phi$ is increasing whenever f is decreasing
• C. $\phi$ is decreasing whenever f is decreasing
• D. $\phi$ is decreasing if $f^{'}(x) = -11$

Answer 1

Answer: (A), (D)

$\phi$ is increasing whenever f is increasing, $\phi$ is decreasing if $f^{'}(x) = -11$

Answer 2

Let $\phi(x) = (f(x))^3-3(f(x))^2 + 4f(x) + 5x + 3 \sin x + 4 \cos x$.
To determine when $\phi(x)$ is increasing or decreasing, we need to find its derivative $\phi'(x)$.
Using the chain rule and differentiation rules, we get:
$\phi'(x) = \frac{d}{dx}((f(x))^3) - 3\frac{d}{dx}((f(x))^2) + 4\frac{d}{dx}(f(x)) + \frac{d}{dx}(5x) + \frac{d}{dx}(3 \sin x) + \frac{d}{dx}(4 \cos x)$
$\phi'(x) = 3(f(x))^2 f'(x) - 3(2f(x) f'(x)) + 4f'(x) + 5 + 3 \cos x - 4 \sin x$
$\phi'(x) = 3(f(x))^2 f'(x) - 6f(x) f'(x) + 4f'(x) + 5 + 3 \cos x - 4 \sin x$
Factor out $f'(x)$ from the first three terms:
$\phi'(x) = f'(x) (3(f(x))^2 - 6f(x) + 4) + 5 + 3 \cos x - 4 \sin x$

Let $y = f(x)$. Consider the quadratic expression $3y^2 - 6y + 4$. We can complete the square:
$3y^2 - 6y + 4 = 3(y^2 - 2y) + 4 = 3(y^2 - 2y + 1 - 1) + 4 = 3((y-1)^2 - 1) + 4 = 3(y-1)^2 - 3 + 4 = 3(y-1)^2 + 1$.
Since $(y-1)^2 \ge 0$ for all real $y$, $3(y-1)^2 \ge 0$, so $3(y-1)^2 + 1 \ge 1$.
Thus, $3(f(x))^2 - 6f(x) + 4 \ge 1$ for all $x \in R$.

Now consider the term $5 + 3 \cos x - 4 \sin x$. The expression $3 \cos x - 4 \sin x$ can be written in the form $R \cos(x+\alpha)$, where $R = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = 5$.
So, $3 \cos x - 4 \sin x = 5 (\frac{3}{5} \cos x - \frac{4}{5} \sin x)$. Let $\cos \alpha = 3/5$ and $\sin \alpha = 4/5$. Then $3 \cos x - 4 \sin x = 5 (\cos \alpha \cos x - \sin \alpha \sin x) = 5 \cos(x+\alpha)$.
The range of $\cos(x+\alpha)$ is $[-1, 1]$, so the range of $5 \cos(x+\alpha)$ is $[-5, 5]$.
The term $5 + 3 \cos x - 4 \sin x = 5 + 5 \cos(x+\alpha)$. Its range is $[5-5, 5+5] = [0, 10]$.
Thus, $5 + 3 \cos x - 4 \sin x \ge 0$ for all $x \in R$.

So, $\phi'(x) = f'(x) (3(f(x))^2 - 6f(x) + 4) + (5 + 3 \cos x - 4 \sin x)$.
Let $A(x) = 3(f(x))^2 - 6f(x) + 4$ and $B(x) = 5 + 3 \cos x - 4 \sin x$.
We have $A(x) \ge 1$ and $B(x) \ge 0$.
$\phi'(x) = f'(x) A(x) + B(x)$.

Let's examine the given options:

1. $\phi$ is increasing whenever f is increasing.
   If $f$ is increasing, then $f'(x) \ge 0$.
   Since $f'(x) \ge 0$ and $A(x) \ge 1$, the term $f'(x) A(x) \ge 0$.
   Since $B(x) \ge 0$, $\phi'(x) = f'(x) A(x) + B(x) \ge 0 + 0 = 0$.
   If $f'(x) > 0$ for an interval, then $f'(x)A(x) > 0$ (since $A(x) \ge 1$), and $B(x) \ge 0$, so $\phi'(x) > 0$.
   If $f'(x) = 0$ for an interval, then $\phi'(x) = B(x) \ge 0$.
   In both cases, $\phi'(x) \ge 0$. Thus, $\phi$ is increasing whenever $f$ is increasing. This option is correct.

2. $\phi$ is increasing whenever f is decreasing.
   If $f$ is decreasing, then $f'(x) \le 0$.
   $\phi'(x) = f'(x) A(x) + B(x)$. Since $f'(x) \le 0$ and $A(x) \ge 1$, $f'(x) A(x) \le 0$.
   So $\phi'(x) = (\le 0) + (\ge 0)$. The sign of $\phi'(x)$ is not guaranteed to be positive. For example, if $f'(x)$ is a large negative number, $f'(x)A(x)$ will be a large negative number, which might outweigh $B(x)$. This option is incorrect.

3. $\phi$ is decreasing whenever f is decreasing.
   If $f$ is decreasing, then $f'(x) \le 0$.
   $\phi'(x) = f'(x) A(x) + B(x) = (\le 0) + (\ge 0)$.
   The sign of $\phi'(x)$ is not guaranteed to be negative. For example, if $f'(x) = 0$, $\phi'(x) = B(x) \ge 0$, so $\phi$ is increasing or constant. If $f'(x)$ is slightly negative, $f'(x)A(x)$ might be small in magnitude and negative, while $B(x)$ can be up to 10, making $\phi'(x)$ positive. This option is incorrect.

4. $\phi$ is decreasing if $f^{'}(x) = -11$.
   If $f'(x) = -11$, then $\phi'(x) = (-11) A(x) + B(x)$.
   We know $A(x) \ge 1$ and $B(x) \in [0, 10]$.
   So, $-11 A(x) \le -11(1) = -11$.
   $\phi'(x) = -11 A(x) + B(x)$. The maximum value of $B(x)$ is 10.
   Therefore, $\phi'(x) \le -11 + 10 = -1$.
   Since $\phi'(x) \le -1 < 0$ for all $x$, $\phi(x)$ is strictly decreasing if $f'(x) = -11$. This option is correct.

The question asks for the correct option(s). Both option 1 and option 4 are correct.