Question

F(x) =√16-x^2 +9/x-3

Answer 1

Ответ:на фото.

Объяснение:может я неправильно поняла,но надеюсь,что помогла и хоть что-то понятно с моего почерка)

Answer 2

To analyze the function $f(x) = \sqrt{16 - x^2} + \frac{9}{x - 3}$, we'll first consider its domain, then look at its behavior, and finally find any critical points.

Domain:

For the square root function to be defined, the expression inside it must be non-negative. This gives us:

$16 - x^2 \geq 0$

Solving for $x$, we get:

$x^2 \leq 16$

Taking the square root of both sides, we have:

$|x| \leq 4$

So the domain of the function is $-4 \leq x \leq 4$, with $x \neq 3$ (since $x$ cannot be 3 due to the denominator in the second term).

Behavior:

1. As $x$ approaches 4 or -4 from within the domain, the square root term approaches zero. This is because as $x$ becomes very close to 4 or -4, $16 - x^2$ gets very close to zero, and the square root of zero is zero.

2. As $x$ approaches 3 from either side (within the domain), the second term approaches infinity. This is because as $x$ gets very close to 3, $x - 3$ gets very close to zero, and the reciprocal of a very small number is a very large number.

Critical Points:

To find the critical points, we'll take the derivative of $f(x)$ and set it equal to zero.

$f(x) = \sqrt{16 - x^2} + \frac{9}{x - 3}$

Taking the derivative:

$f'(x) = -\frac{x}{\sqrt{16 - x^2}} - \frac{9}{(x - 3)^2}$

Setting $f'(x) = 0$ and solving for $x$:

$-\frac{x}{\sqrt{16 - x^2}} - \frac{9}{(x - 3)^2} = 0$

This equation does not have any real solutions within the given domain.

Summary:

• Domain: $-4 \leq x \leq 4$ with $x \neq 3$
• Behavior:
  • As $x$ approaches 4 or -4, $f(x)$ approaches 0.
  • As $x$ approaches 3 from either side, $f(x)$ approaches positive infinity.

Please note that this analysis is based on the given domain. If the domain were different, the behavior of the function might change accordingly.