Show that the equation x^3-3x^2+3=0 has a solution between x=2 and x=3
1 answer:
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This is t<span>he </span>Greatest Integer Function,<span> denoted by </span>![y = [x]](https://tex.z-dn.net/?f=y%20%3D%20%5Bx%5D)
<span>. For all real numbers, x, <em>the </em></span><em>greatest integer function</em><span><em> returns the largest </em></span><em>integer</em><span><em> less than or equal to x</em>. In essence, it rounds down a real number to the nearest </span>integer<span>.
a. FALSE. Because f(4.9) will round down to 4.
b. TRUE.
c. no it's not the absolute value function. that has V-shape
d. TRUE.
e. FALSE. because for every unique x-value you don't always get a unique y value. .... f(5)=5, f(5.1)=5, f(5.76)=5, etc.</span>
a. FALSE. Because f(4.9) will round down to 4.
b. TRUE.
c. no it's not the absolute value function. that has V-shape
d. TRUE.
e. FALSE. because for every unique x-value you don't always get a unique y value. .... f(5)=5, f(5.1)=5, f(5.76)=5, etc.</span>
Answer:
27) x = 2^(y) – 5.
Asymptote: x = -5.
D: x > -5; (-5, infinity).
R: -infinity < f(x) < infinity; ARN;
(-infinity, infinity).
x → -infinity, f(x) → -infinity.
x → +infinity, f(x) → +infinity.
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28) x = 2^-(y–3).
Asymptote: x = 0.
D: x > 0; (0, infinity).
R: -infinity < f(x) < infinity; ARN;
(-infinity, infinity).
x → -infinity, f(x) → +infinity.
x → +infinity, f(x) → -infinity.
________________________
29) x = 4^(y–2) + 1.
Asymptote: x = 1.
D: x > 1; (1, infinity).
R: -infinity < f(x) < infinity; ARN;
(-infinity, infinity).
x → -infinity, f(x) → -infinity.
x → +infinity, f(x) → +infinity.
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