Finding domain and range (1 Viewer)

Flop21 · Oct 8, 2015

You need to know that f(x) is the function, f'(x) shows you if it's increasing or decreasing, f'(x) tells you the concavity (thus you can work out if it's a min or max from concavity).

increasing >0

decreasing <0

so find values that f'(x) > 0 and use the graph to help you, increasing means the gradient is positive / sloping upwards, going upwards from left to right.

BlueGas · Oct 8, 2015

BlueGas said:
Okay so to find the x values when the function is increasing, I look at the first derivative when it's positive, f'(x) > 0, that's a minimum, so I look at the minimum on the graph?

Woops I confused the first and second derivative.

So I get that after -2 the graph is increasing and the same thing for 1 but I don't know how to put the answer in the proper form.

spatula232 · Oct 8, 2015

BlueGas said:
Woops I confused the first and second derivative.

So I get that after -2 the graph is increasing and the same thing for 1 but I don't know how to put the answer in the proper form.

-2 < x < 0 and x > 1

InteGrand · Oct 8, 2015

BlueGas said:
Okay so here's an example of a question, when they say "for what values is the x function increasing", does that mean what's the domain?

Also what would the domain and range be for this example?

$It's a polynomial function so its domain is $\mathbb{R}$ (i.e. the set of all real $x$). We can see from the graph that the range is the set of all real $y$ such that $y\geq -32$.$

InteGrand · Oct 8, 2015

spatula232 said:
Function is increasing where the first derivation is positive, i.e. > 0

That would mean -2 < x < 0 and x > 1 (Assuming your graph is correct)

It can't be Greater than or equal to because at that turning point f'(x)=0 and therefore not >0 (as function is increasing at where f'(x)>0)

$Actually, we should include equality cases in those, that is, the function is increasing for all $x$ such that $-2\leq x \leq 0$ or $x \geq 1$. This is because the definition of a function to be \textsl{increasing} on an interval $I\subseteq\mathbb{R}$ is that $f(x_2) \geq f(x_1)$ for all $x_1,x_2 \in I$ such that $x_2\geq x_1$. We can see that this is the case if we include the equality signs on the intervals you gave, so we need to include them to get the maximal intervals where $f$ is increasing.$

InteGrand · Oct 8, 2015

BlueGas · Oct 8, 2015

spatula232 said:
-2 < x < 0 and x > 1

Can't it also be written as x > -2 and x > 1?

InteGrand · Oct 8, 2015

BlueGas said:
Can't it also be written as x > -2 and x > 1?

$No, because there are some places where $x>-2$ but the function is not increasing. E.g. for all $x$ in the interval $0\leq x \leq 1$, we have $x>-2$, yet the function is not increasing on this interval.$

BlueGas · Oct 8, 2015

InteGrand said:
$No, because there are some places where $x>-2$ but the function is not increasing. E.g. for all $x$ in the interval $0\leq x \leq 1$, we have $x>-2$, yet the function is not increasing on this interval.$

For example at 0,0 it's not increasing right?

InteGrand · Oct 8, 2015

BlueGas said:
For example at 0,0 it's not increasing right?

Technically it is increasing. The term "increasing" in maths includes cases where the function is constant. If we want to avoid these cases, we need to use the term strictly increasing. Also technically these terms refer to intervals. If the interval is just a single point where the function is defined, it must be increasing there, as it trivially satisfies the definition of increasing in any "interval" of just one point. It also trivially satisfies the definition of decreasing on any subset of the reals that contains only one point. Such an interval consisting of only one point is called a degenerate interval.

BlueGas · Oct 8, 2015

Okay so how would I find the domain and range for this? (Even though it only asks for domain but I want to know how to find the domain and the range)

rand_althor · Oct 8, 2015

BlueGas said:
Okay so how would I find the domain and range for this? (Even though it only asks for domain but I want to know how to find the domain and the range)

The normal domain for ln(x) is x>0. In this case, we have ln[tan(x)], so we require tan(x)>0 over the domain 0≤x≤2π. Might be easier to find for what values of x this inequality holds true if you draw a graph of tan(x).

InteGrand · Oct 8, 2015

BlueGas said:
Okay so how would I find the domain and range for this? (Even though it only asks for domain but I want to know how to find the domain and the range)

$For the range, note that the tan function attains every real value as $x$ varies over $[0,2\pi]$ (this notation means the closed interval from 0 to $2\pi$), and in particular attains every positive value, so the $\ln (\tan x)$ function will attain every real value, because we know that as $u$ varies over the positive reals, $\ln u$ varies over the reals. So the range of $f(x)=\ln (\tan x)$ is $\mathbb{R}$.$

Finding domain and range (1 Viewer)

Flop21

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