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3 years ago

Write the equation of the function g(x) if g(x) = f(x – 3) +1 and f(x) = x3 + 1.

1 answer:

4 0

We can find the value of f (x-3) by putting x-3 in place of x in f (x)
therefore,
f (x-3)=(x-3)3 + 1
by putting the value of f (x-3) in g (x)
g (x)= (x-3)3 + 1 +1
= (x-3)3 + 2

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Answer:

27405

Step-by-step explanation:

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3 years ago

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3 years ago

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Answer:

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3 years ago

If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .

Tems11 [23]

Divide through everything by b :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since a/b < c/d, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by b/d to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (a + c)/(b + d) < c/d.

For the other side, you can do something similar and divide through everything by d :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and a/b < c/d tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

and so (a + c)/(b + d) > a/b.

Then together we get the desired inequality.

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3 years ago

Find the domain and range of the graph

ozzi

9514 1404 393

Answer:

domain: [-6, -1)
range: [-5, 4]

Step-by-step explanation:

You have correctly determined the domain to be from x = -6 to x < -1. In interval notation, that would be ...

[-6, -1)

The range is the vertical extent of the graph. It extends from a minimum of y = -5 to a maximum of y = +4 at the top of the curve. In interval notation, that would be ...

[-5, 4]

8 0

2 years ago