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11 months ago

Given the values of the linear functions f (x) and g(x) in the tables, where is (f – g)(x) positive?

Mathematics

1 answer:

alexandr402 [8]11 months ago

4 0

Comparing the functions, from the tables, it is found that (f - g)(x) is positive in the interval (–∞, 9) .

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For the subtraction function, we simply subtract both functions, thus:

$(f - g)(x) = f(x) - g(x)$

It is positive if f is greater than g, that is: f(x) > g(x).
It is a linear function, so one function is greater before the equality, one after.
They are equal at x = 9.
If x < 9, f(x) > g(x), and thus, (f - g)(x) is positive, which means that the desired interval is:

(–∞, 9)

A similar problem is given at brainly.com/question/24610273

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C(q) = -(1/4)*q^3 + 3q^2 - 12q + OH

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1 year ago

se the function to show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0). f(x, y) = 9x2y x4 + y

alexandr1967 [171]

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It is proved that $f_x, f_y$ exixts at (0,0) but not differentiable there.

Step-by-step explanation:

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$f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)$

To show exixtance of $f_x(0,0), f_y(0,0)$ we take,

$f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0$ exists.

And,

$f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}$ exists.

To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,

$\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}$ where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).

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