Someone please answer this question

Mathematics

1 answer:

gavmur [86]3 years ago

7 0

Answer:

$f$ is a function because each student, or each element of the domain of the function corresponds to one element of the range.

Step-by-step explanation:

Here are some of the important factors for a relationship to be a function.

As we know that a function is a relation between sets that associates to every element of a first set exactly one element of the second set.

In the giving diagram of the question, it is clearly observable that:

Each student in Mr. Well's history class associates to one Final exam score.

Therefore, $f$ is a function.

We know that the domain of a function is the set of all the x-values of an ordered pair of a set for which the function is defined.

Here,

The domain of $f$ is : {Student}

And range of a function is the set of all the y-values of a ordered pair of a set.

So,

The range of $f$ is : {Final Exam Score}

Hence, we conclude that $f$ is a function because each student, or each element of the domain of the function corresponds to one element of the range.

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Translate this phrase into an algebraic expression.

alexira [117]

Answer:

5+(11 x n)

Step-by-step explanation:

I hope this helps

6 0

3 years ago

2. Find the stationary points for the function

vovikov84 [41]

Answer & step-by-step explanation:

Stationary points are the points where the first derivative is equal to zero.

Let's calculate it using the power rule (exponent comes forward, decrease exponent by 1) and the fact that the derivative is a linear operation (that is $D[a\ f(x) + b\ g(x)] = a Df(x) + b Dg(x)$ )

The first derivative is then

$y' = \frac13 (3x^2) - \frac12 (2x) -6 = x^2-x-6 = (x+2)(x-3)$

Note that the last passage is not strictly needed, but it's really helpful to find stationary points, when in this next passage we set it equal to zero. Alternatively, you can use the quadratic formula if you can pull the factors out of your head right away.

$y'=0 \rightarrow (x+2)(x-3) = 0 \implies x=-2 || x=3$

These two point could be maxima, minima, or inflection points. To check them you can either see how the sign of the first derivative goes, or check the sign of the second derivative, as you're required.

The rules states that if the second derivative evaluated in that point is negative we have a maximum, if it's positive we have a minimum, and if we have a zero we keep derivating until we get a non-zero value.

In our case, the second derivative we get by calculating the derivative again and we get $y'' = 2x-1$ . Evaluating it at both points we get