0.2 Functions and Graphs Hàm số và đồ thị

(a) $A = \pi\left(\dfrac{d}{2}\right)^2 = \dfrac{\pi d^2}{4}.$ (b) With $d = 6$: $A = \dfrac{\pi \cdot 6^2}{4} = 9\pi \approx 28.27$ cm$^2.$

(a) $h = \dfrac{\sqrt{3}}{2}s,$ from the Pythagorean theorem on half the triangle. (b) With $s = 6$: $h = 3\sqrt{3} \approx 5.20$ m.

(a) A cube has $6$ square faces of area $e^2,$ so $S = 6e^2.$ (b) With $e = 4$: $S = 6 \cdot 4^2 = 96$ ft$^2.$

(a) $V = \dfrac{4}{3}\pi r^3.$ (b) With $r = 2$: $V = \dfrac{4}{3}\pi \cdot 8 = \dfrac{32\pi}{3} \approx 33.51$ cm$^3.$

(a) Every real $x$ can be squared, so the domain is all real numbers. The largest value is $5$ at $x = 0,$ so the range is $(-\infty, 5].$ (b) Sketch a downward parabola with vertex $(0, 5).$

(a) Every real $x$ works, so the domain is all real numbers. Since $x^2 \ge 0,$ $y \ge -4,$ so the range is $[-4, \infty).$ (b) Upward parabola with vertex $(0, -4).$

(a) The square root needs $x - 2 \ge 0,$ so the domain is $[2, \infty).$ Since $\sqrt{x-2} \ge 0,$ $y \ge 3,$ so the range is $[3, \infty).$ (b) Start at $(2, 3)$ and rise to the right.

(a) $-x \ge 0,$ so the domain is $(-\infty, 0].$ The square root is $\ge 0,$ so $y \le 0$ and the range is $(-\infty, 0].$ (b) The graph is the lower half of a parabola opening left, from $(0, 0)$ down to the left.

(a) The denominator cannot be $0,$ so $x \ne 3.$ The fraction is never $0,$ so $y \ne 0.$ (b) Two branches: one above and right of $(3, 0),$ one below and left.

(a) A fourth root needs $-x \ge 0,$ so the domain is $(-\infty, 0].$ The root is $\ge 0,$ so the range is $[0, \infty).$ (b) Curve through $(0, 0)$ and $(-1, 1),$ going left and up.

(a) $x \ne 0,$ so the domain is $x \ne 0.$ Since $\dfrac{1}{x} \ne 0,$ $y \ne 2,$ so the range is $y \ne 2.$ (b) Hyperbola shifted up by $2.$

(a) $x \ne 0,$ so the domain is $x \ne 0.$ Since $\dfrac{1}{x^2} > 0,$ $y > 2,$ so the range is $(2, \infty).$ (b) Two branches above $y = 2,$ rising near $x = 0.$

(a) A cube root accepts any real $x,$ and it takes every real value. Domain and range are both $(-\infty, \infty).$ (b) An S-shaped curve through $(0, 0),$ $(1, 1)$ and $(-1, -1).$

(a) $4 - x \ge 0,$ so the domain is $(-\infty, 4].$ The root is $\ge 0,$ so the range is $[0, \infty).$ (b) Start at $(4, 0)$ and rise to the left.

(a) A cube root accepts every real number, so the domain is $(-\infty, \infty).$ The cube root takes all real values, so the range is $(-\infty, \infty).$ (b) Curve through $(0, 1),$ $(1, 0)$ and $(-1, 0).$

(a) $16 - x^2 \ge 0$ gives $-4 \le x \le 4.$ The root is between $0$ and $4,$ so the range is $[0, 4].$ (b) Upper half of the circle of radius $4.$

(a) The fifth root of $x$ exists for every real $x,$ so the domain is $(-\infty, \infty).$ The value $x^{2/5}$ is never negative, so the range is $[0, \infty).$ (b) Curve through $(0, 0),$ $(1, 1)$ and $(-1, 1).$

(a) $x^{3/2} = (\sqrt{x})^3$ needs $x \ge 0,$ so the domain is $[0, \infty).$ It is never negative, so the range is $[0, \infty).$ (b) Curve through $(0, 0)$ and $(1, 1),$ rising faster and faster.

(a) A cube root accepts every real number, so the domain and the range are both $(-\infty, \infty).$ (b) The graph of $\sqrt[3]{x}$ moved $2$ units to the left.

(a) We need $9 - x^2 > 0,$ so $-3 < x < 3.$ Since $0 < \sqrt{9 - x^2} \le 3,$ the reciprocal is $\ge \dfrac{1}{3},$ so the range is $[\frac{1}{3}, \infty).$ (b) Curve that rises steeply near $x = \pm 3$ and has its lowest point at $(0, \frac{1}{3}).$

$f(-x) = (-x)^6 = x^6 = f(x),$ so the function is even.

$f(-x) = -x + x^2.$ It is neither $f(x)$ nor $-f(x),$ so the function is neither.