Question

Find the set of values of a for which the function,

$f(x) = (1 - \frac{\sqrt{21-4a-a^2}}{a+1})x^3 + 5x + \sqrt{7}$ is increasing at every point of its domain.

Answer 1

[-7, -1) ∪ [2, 3]

Answer 2

To find the set of values of 'a' for which the function $f(x)$ is increasing at every point of its domain, we need to follow these steps:

1. Determine the domain of 'a' for the function to be defined:

The function $f(x) = (1 - \frac{\sqrt{21-4a-a^2}}{a+1})x^3 + 5x + \sqrt{7}$ involves a square root and a fraction.

• For the term $\sqrt{21-4a-a^2}$ to be defined, the expression under the square root must be non-negative: $21-4a-a^2 \ge 0$ $a^2+4a-21 \le 0$

Factorizing the quadratic: $(a+7)(a-3) \le 0$

This inequality holds for $a \in [-7, 3]$.

• For the term $\frac{\sqrt{21-4a-a^2}}{a+1}$ to be defined, the denominator cannot be zero: $a+1 \ne 0 \implies a \ne -1$.

Combining these conditions, the permissible values for 'a' are $a \in [-7, 3]$ excluding $a=-1$. So, $a \in [-7, -1) \cup (-1, 3]$.

2. Calculate the derivative of the function, $f'(x)$:

Let $C = 1 - \frac{\sqrt{21-4a-a^2}}{a+1}$.

Then $f(x) = Cx^3 + 5x + \sqrt{7}$.

Differentiating with respect to $x$: $f'(x) = 3Cx^2 + 5$.

3. Set the condition for $f(x)$ to be increasing:

A function $f(x)$ is increasing at every point of its domain if $f'(x) \ge 0$ for all $x$ in its domain.

So, we need $3Cx^2 + 5 \ge 0$ for all $x \in R$.

This is a quadratic expression in $x$.

• If $C > 0$, then $3Cx^2 \ge 0$ for all $x$. Thus, $3Cx^2 + 5 \ge 5$, which is always positive. In this case, $f(x)$ is strictly increasing.
• If $C = 0$, then $f'(x) = 5$, which is always positive. In this case, $f(x)$ is strictly increasing.
• If $C < 0$, then $3Cx^2$ is always less than or equal to zero. As $|x|$ increases, $3Cx^2$ becomes a large negative number, making $3Cx^2 + 5$ negative for sufficiently large $|x|$. Therefore, $f'(x)$ would not be $\ge 0$ for all $x$.

Thus, for $f'(x) \ge 0$ to hold for all $x$, we must have $C \ge 0$.

4. Solve the inequality $C \ge 0$ for 'a':

$1 - \frac{\sqrt{21-4a-a^2}}{a+1} \ge 0$

$\frac{\sqrt{21-4a-a^2}}{a+1} \le 1$

We need to solve this inequality considering the domain for 'a' found in step 1 ($a \in [-7, -1) \cup (-1, 3]$). We analyze two cases based on the sign of the denominator $(a+1)$:

Case 1: $a+1 > 0 \implies a > -1$.

Considering the domain for 'a', this case applies for $a \in (-1, 3]$.

Since $a+1$ is positive, we can multiply both sides by $(a+1)$ without changing the inequality direction:

$\sqrt{21-4a-a^2} \le a+1$

Both sides are non-negative (LHS is a square root, RHS is positive for $a > -1$). So, we can square both sides:

$21-4a-a^2 \le (a+1)^2$

$21-4a-a^2 \le a^2+2a+1$

$0 \le a^2+2a+1 + a^2+4a-21$

$0 \le 2a^2+6a-20$

Dividing by 2:

$0 \le a^2+3a-10$

Factorizing the quadratic:

$0 \le (a+5)(a-2)$

This inequality holds for $a \in (-\infty, -5] \cup [2, \infty)$.

Intersecting this solution with the condition for this case ($a \in (-1, 3]$):

The intersection is $a \in [2, 3]$.

Case 2: $a+1 < 0 \implies a < -1$.

Considering the domain for 'a', this case applies for $a \in [-7, -1)$.

When we multiply both sides of $\frac{\sqrt{21-4a-a^2}}{a+1} \le 1$ by $(a+1)$ (which is negative), we must reverse the inequality sign:

$\sqrt{21-4a-a^2} \ge a+1$

In this case, the left-hand side ($\sqrt{21-4a-a^2}$) is always non-negative (as it's a square root). The right-hand side ($a+1$) is strictly negative.

A non-negative number is always greater than or equal to a negative number. Therefore, this inequality is always true for all 'a' where the square root is defined and $a+1 < 0$.

Intersecting this with the condition for this case ($a \in [-7, -1)$):

The solution for this case is $a \in [-7, -1)$.

5. Combine the solutions from both cases:

The set of values of 'a' for which $f(x)$ is increasing at every point of its domain is the union of the solutions from Case 1 and Case 2:

$a \in [-7, -1) \cup [2, 3]$.