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Morgarella [4.7K]
3 years ago
7

For each function, do the following: a) state the name of the function, b) identify the independent and dependent variable, c) i

dentify the rule that assigns exactly one output to the very input, and d) evaluate the function for the given input value.

Mathematics
1 answer:
user100 [1]3 years ago
5 0
<h2>Hello</h2>

The answers are:

a) The name of the function is y, and it's a Square Root Function.

b) Independent variable : x , dependent variable : y

c) The rule that assigns exactly one output to the very input is called "function".

d) y=\sqrt{25}=5

<h2>Why?</h2>

The name of a function can be also given as a single letter (y). For this exercise,  the name of the function is "y", and it's a Square Root Function.

The independent variable of a function is the variable we assign the different values. For this exercise, the independent variable is designated with the letter "x".

The dependent variable is the function itself (y),  and it's called "dependent" variable because its values will always depend on the "independent variable".

A function is the rule that states that there is exactly one output (range value) to the very input (domain value).  A function only exists when there is exactly one output value (range) for each given input (domain), if there is more than one output for each input, the function does not exist.

To evaluate a function we assign values to the independent variable(x), hence:

y=\sqrt{25}=5

Have a nice day!

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Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.
Leni [432]

Answer:

a)

g(2.9) \approx -6.6

g(3.1) \approx -3.4

b)

The values are too small since g'' is positive for both values of x in. I'm speaking of the x values, 2.9 and 3.1.

Step-by-step explanation:

a)

The point-slope of a line is:

y-y_1=m(x-x_1)

where m is the slope and (x_1,y_1) is a point on that line.

We want to find the equation of the tangent line of the curve g at the point (3,-5) on g.

So we know (x_1,y_1)=(3,-5).

To find m, we must calculate the derivative of g at x=3:

m=g'(3)=(3)^2+7=9+7=16.

So the equation of the tangent line to curve g at (3,-5) is:

y-(-5)=16(x-3).

I'm going to solve this for y.

y-(-5)=16(x-3)

y+5=16(x-3)

Subtract 5 on both sides:

y=16(x-3)-5

What this means is for values x near x=3 is that:

g(x) \approx 16(x-3)-5.

Let's evaluate this approximation function for g(2.9).

g(2.9) \approx 16(2.9-3)-5

g(2.9) \approx 16(-.1)-5

g(2.9) \approx -1.6-5

g(2.9) \approx -6.6

Let's evaluate this approximation function for g(3.1).

g(3.1) \approx 16(3.1-3)-5

g(3.1) \approx 16(.1)-5

g(3.1) \approx 1.6-5

g(3.1) \approx -3.4

b) To determine if these are over approximations or under approximations I will require the second derivative.

If g'' is positive, then it leads to underestimation (since the curve is concave up at that number).

If g'' is negative, then it leads to overestimation (since the curve is concave down at that number).

g'(x)=x^2+7

g''(x)=2x+0

g''(x)=2x

2x is positive for x>0.

2x is negative for x.

That is, g''(2.9)>0 \text{ and } g''(3.1)>0.

So 2x is positive for both values of x which means that the values we found in part (a) are underestimations.

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