Given that f(t) = 1 − 3/t and the range is −1, 3, 7 find the domain.

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Part b) To find the domain D such that the range of f(x) = (1)/(x−2) is [−1, 0). Step 1: Solve y = (1)/(x−2) for x. y = (1)/(x−2); x − 2 = 1/y; x = 2 + 1/y, y ≠ 0; Step 2: f'(x) = −(1)/((x−2)²) < 0, so f is strictly decreasing on (−∞, 2) and on (2, ∞). On (−∞, 2), f(x) < 0: as x → −∞, f(x) → 0^−; as x → 2^−, f(x) → −∞. Step 3: Solve f(x) = −1. (1)/(x−2) = −1; x−2 = −1; x = 1; f(1) = −1. Step 4: Since f is strictly decreasing on (−∞, 2), on (−∞, 1], f decreases from 0^− to −1. Thus, f((−∞, 1]) = [−1, 0). Domain: (−∞, 1] (-∞,1] Part c) f(x) = x² − 2x. Show f(x) ≥ −1 for all real x. Step 1: Complete the square. f(x) = x² − 2x = (x² − 2x + 1) − 1 = (x−1)² − 1 Step 2: (x−1)² ≥ 0 for all real x. f(x) = (x−1)² − 1 ≥ 0 − 1 = −1; Equality at x=1: f(1) = −1. Alternatively, f'(x) = 2x − 2 = 0 gives x=1, f''(x)=2>0 so minimum f(1)=−1. Thus, f(x) ≥ −1 for all x. f(x)≥−1 Part d) Given aⁿ = bⁿ, show ( a/b )ⁿ = 1 (assume b ≠ 0). Step 1: Divide both sides by bⁿ. aⁿ = bⁿ; (aⁿ)/(bⁿ) = (bⁿ)/(bⁿ); ( a/b )ⁿ = 1; If b=0, then aⁿ=0 so a=0, but a/b undefined. **( a/b )ⁿ = 1

Given that f(t) = 1 − (3)/(t) and the range is \-1, 3, 7\, find the domain.

Answers

Step-by-step solution

Part b· 4 steps

Solve $y = \frac{1}{x-2}$ for $x$ .

$f'(x) = -\frac{1}{(x-2)^2} < 0$ , so $f$ is strictly decreasing on $(-\infty, 2)$ and on $(2, \infty)$ .

Solve $f(x) = -1$ .

Since $f$ is strictly decreasing on $(-\infty, 2)$ , on $(-\infty, 1]$ , $f$ decreases from $0^-$ to $-1$ .

Part c· 2 steps
$f(x)\ge-1$

Complete the square.

$(x-1)^2 \ge 0$ for all real $x$ .

Part d· 1 step
$\left( \frac{a}{b} \right)^{n} = 1$

Divide both sides by $b^n$ .

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More Mathematics Questions

Given that f(t) = 1 − (3)/(t) and the range is \-1, 3, 7\, find the domain.

Answers

Step-by-step solution

Part b· 4 steps

Solve y=1x−2y = \frac{1}{x-2}y=x−21​ for xxx.

f′(x)=−1(x−2)2<0f'(x) = -\frac{1}{(x-2)^2} < 0f′(x)=−(x−2)21​<0, so fff is strictly decreasing on (−∞,2)(-\infty, 2)(−∞,2) and on (2,∞)(2, \infty)(2,∞).

Solve f(x)=−1f(x) = -1f(x)=−1.

Since fff is strictly decreasing on (−∞,2)(-\infty, 2)(−∞,2), on (−∞,1](-\infty, 1](−∞,1], fff decreases from 0−0^-0− to −1-1−1.

Part c· 2 stepsf(x)≥−1f(x)\ge-1f(x)≥−1