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2 years ago

Help !!!See question in image.Please show workings .

Mathematics

1 answer:

ivolga24 [154]2 years ago

5 0

Answer:

see explanation

Step-by-step explanation:

Given f(x) then the derivative f'(x) is

f'(x) = lim(h tends to 0 ) $\frac{f(x+h)-f(x)}{h}$

= lim ( h to 0 ) $\frac{4(x+h)^2-2-(4x^2-2)}{h}$

= lim ( h to 0 ) $\frac{4(x+h)^2-2-4x^2+2}{h}$

= lim( h to 0 ) $\frac{4x^2+8hx+4h^2-4x^2}{h}$

= lim( h to 0 ) $\frac{8hx+4h^2}{h}$

= lim ( h to 0 ) $\frac{4h(2x+h)}{h}$ ← cancel h on numerator/ denominator

= lim ( h to 0 ) 4(2x + h) ← let h go to zero

f'(x) = 8x

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9 months ago

Tell whether the ordered pair is a solution of the system of linear equations.

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First thing you would do is plug in (-1) to x, and 6 to y.

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For #1, it would be no solution because the numbers do not equal each other. (Fun fact) If one of them ends up as a solution, but the other is no solution, the answer would be no solutions because BOTH of them have to be a solution to the equations given in order to be a solution.

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2 years ago

For the function g(x)=3-8(1/4)^2-x

Reika [66]

Using function concepts, it is found that:

a) The y-intercept is y = 2.5.

b) The horizontal asymptote is x = 3.

c) The function is decreasing.

d) The domain is $(-\infty,\infty)$ and the range is $(-\infty,3)$ .
e) The graph is given at the end of the answer.

------------------------------------

The given function is:

$g(x) = 3 - 8\left(\frac{1}{4}\right)^{2-x}$

------------------------------------

Question a:

The y-intercept is g(0), thus:

$g(0) = 3 - 8\left(\frac{1}{4}\right)^{2-0} = 3 - 8\left(\frac{1}{4}\right)^{2} = 3 - \frac{8}{16} = 3 - 0.5 = 2.5$

The y-intercept is y = 2.5.

------------------------------------

Question b:

The horizontal asymptote is the limit of the function when x goes to infinity, if it exists.

$\lim_{x \rightarrow -\infty} g(x) = \lim_{x \rightarrow -\infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2+\infty} = 3 - 8\left(\frac{1}{4}\right)^{\infty} = 3 - 8\frac{1^{\infty}}{4^{\infty}} = 3 -0 = 3$

--------------------------------------------------

$\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2-\infty} = 3 - 8\left(\frac{1}{4}\right)^{-\infty} = 3 - 8\times 4^{\infty} = 3 - \infty = -\infty$

Thus, the horizontal asymptote is x = 3.

--------------------------------------------------

Question c:

The limit of x going to infinity of the function is negative infinity, which means that the function is decreasing.

--------------------------------------------------

Question d:

Exponential function has no restrictions in the domain, so it is all real values, that is $(-\infty,\infty)$ .
From the limits in item c, the range is: $(-\infty,3)$

--------------------------------------------------

The sketching of the graph is given appended at the end of this answer.

A similar problem is given at brainly.com/question/16533631

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3 years ago

Help !!!See question in image.Please show workings .​

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