The problem defines a piecewise function $f(x)$ as follows: $f(x) = x^2 - 6$ if $x < 0$ $f(x) = 10 - x$ if $x \ge 0$ The problem asks us to: a) Sketch the graph of $f(x)$. b) Find the value(s) of $a$ such that $f(a) = 43$. c) Find the values of $x$ in the domain such that $f(x) = x$.

1. Problem Description

The problem defines a piecewise function $f(x)$ as follows:

$f(x) = x^2 - 6$ if $x < 0$

$f(x) = 10 - x$ if $x \ge 0$

The problem asks us to:

a) Sketch the graph of $f(x)$.

b) Find the value(s) of $a$ such that $f(a) = 43$.

c) Find the values of $x$ in the domain such that $f(x) = x$.

2. Solution Steps

a) Sketch $f(x)$:

For $x < 0$, $f(x) = x^2 - 6$. This is a parabola opening upwards with vertex at $(0, -6)$. Since $x < 0$, we only consider the left half of the parabola.

For $x \ge 0$, $f(x) = 10 - x$. This is a straight line with slope $-1$ and y-intercept $10$. Since $x \ge 0$, we only consider the part of the line to the right of the y-axis.

b) Find $a$ such that $f(a) = 43$:

Case 1: $a < 0$. Then $f(a) = a^2 - 6 = 43$.

$a^2 = 49$

$a = \pm 7$

Since $a < 0$, we have $a = -7$.

Case 2: $a \ge 0$. Then $f(a) = 10 - a = 43$.

$a = 10 - 43 = -33$

Since $a \ge 0$, this case gives no solution.

Thus, the only solution is $a = -7$.

c) Find the values of $x$ such that $f(x) = x$:

Case 1: $x < 0$. Then $f(x) = x^2 - 6 = x$.

$x^2 - x - 6 = 0$

$(x - 3)(x + 2) = 0$

$x = 3$ or $x = -2$.

Since $x < 0$, we have $x = -2$.

Case 2: $x \ge 0$. Then $f(x) = 10 - x = x$.

$2x = 10$

$x = 5$

Since $x \ge 0$, we have $x = 5$.

Therefore, the values of $x$ such that $f(x) = x$ are $x = -2$ and $x = 5$.

3. Final Answer

a) The sketch of $f(x)$ is a parabola $x^2 - 6$ for $x < 0$ and a line $10 - x$ for $x \ge 0$.

b) $a = -7$

c) $x = -2, 5$