Few questions (1 Viewer)

BlueGas · Oct 9, 2015

How do I do these questions? Sorry for so many questions at once.

InteGrand · Oct 9, 2015

BlueGas said:
How do I do these questions? Sorry for so many questions at once.

For Q. (d), use f(x) = |x – 1|.

InteGrand · Oct 9, 2015

BlueGas said:
How do I do these questions? Sorry for so many questions at once.

$For Q.7, $\log _{10} \left(\frac{1}{70} \right)=-\log_{10}70$$

$=-\log_{10}(7\times 10)$

$=-(\log_{10}7 + \log_{10} 10)$

$$=-(a + 1)$, and so the answer is (A).$

InteGrand · Oct 9, 2015

BlueGas said:
How do I do these questions? Sorry for so many questions at once.

$For Question 5, note that since $\cos x=\frac{1}{2}$, we have $\cos^2 x = \frac{1}{4}$, and $\sin x = \sqrt{1-\left( \frac{1}{2}\right)^2}=\frac{\sqrt{3}}{2}>\frac{1}{4}$, so the answer can't be (A). (Took the positive square root since $x$ is acute, so $\sin x>0$.)$

$Clearly, $\tan x = \frac{\sin x}{\cos x}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}>1>\frac{\sqrt{3}}{2}$, and so $\tan x$ is greatest, i.e. the answer is (C).$

BlueGas · Oct 9, 2015

InteGrand said:
For Q. (d), use f(x) = |x – 1|.

The problem is I don't understand sketching that graph.

InteGrand · Oct 9, 2015

BlueGas said:
How do I do these questions? Sorry for so many questions at once.

$For Question (b), note that the shaded segment has area $A=\frac{1}{2}r^2 \left( \theta - \sin \theta \right)$. We are given that this is half the area of the semicircle, so $A=\frac{1}{2}\times \frac{\pi r^2}{2}$. Therefore, comparing with our previous expression for $A$ gives us $\theta - \sin \theta =\pi$.$

InteGrand · Oct 9, 2015

BlueGas said:
The problem is I don't understand sketching that graph.

This is simply the graph of y = |x| shifted right by 1 unit (recall how to sketch transformations of known graphs: http://community.boredofstudies.org/12/mathematics/342876/how-graph-trig-functions.html#post7037599 )

BlueGas · Oct 9, 2015

InteGrand said:
This is simply the graph of y = |x| shifted right by 1 unit (recall how to sketch transformations of known graphs: http://community.boredofstudies.org/12/mathematics/342876/how-graph-trig-functions.html#post7037599 )

But how do I graph |x|? The absolutes confuse me.

Flop21 · Oct 9, 2015

BlueGas said:
But how do I graph |x|? The absolutes confuse me.

Make a table of values, I've concluded that's the easiest way to not stuff up the absolute equations / functions.

kawaiipotato · Oct 9, 2015

BlueGas said:
But how do I graph |x|? The absolutes confuse me.

Draw the y = x curve. Whatever dips under the x-axis, reflect about the x-axis.
.
Generalise: y = |f(x)|, wherever f(x) goes under the x-axis, reflect about the x-axis.

InteGrand · Oct 9, 2015

BlueGas said:
But how do I graph |x|? The absolutes confuse me.

That's a curve I think you should be expected to know.

BlueGas · Oct 10, 2015

InteGrand said:
For Q. (d), use f(x) = |x – 1|.

Also how would I know that the function is |x-1|? Why can't it be for example |x+1|?

InteGrand · Oct 10, 2015

BlueGas said:
Also how would I know that the function is |x-1|? Why can't it be for example |x+1|?

$The function we want is one whose slope is: -1 for all $x<1$, $1$ for all $x>1$, and undefined at $x=1$. You are expected to be familiar with the graph of $y=|x|$, and should notice that this graph has all these properties except that they occur about $x=0$ rather than $x=1$. So we need to just shift the graph of $y=|x|$ to the right by one. If you know your graph transformations, you'll know that this is achieved by replacing $x$ in $y=|x|$ by $x-1$. So the graph we want is that of $y=|x-1|$.$

BlueGas · Oct 10, 2015

Also what would the domain and range be for y = |x - 1|?

If there was also another question asking, for what values is the f(x) function increasing? What would the answer be to that?

davidgoes4wce · Oct 10, 2015

BlueGas said:
Also what would the domain and range be for y = |x - 1|?

If there was also another question asking, for what values is the f(x) function increasing? What would the answer be to that?

Domain would be all real values of x, Range would be y greater than equal to 0

rand_althor · Oct 10, 2015

BlueGas said:
Also what would the domain and range be for y = |x - 1|?

If there was also another question asking, for what values is the f(x) function increasing? What would the answer be to that?

If you draw a graph you should easily be able to figure out both the domain and range. If you don't have one here you go:

Domain: what possible x-values can be subbed in to the function?
Range: what possible y-values are outputted once these x-values are subbed in?

Now to know where f(x) is increasing, check a graph of f'(x) (you posted one in your original post). The function is increasing when f'(x)>0, and decreasing when f'(x)<0. For what x-values do these statements hold true in the graph of f'(x)?

InteGrand · Oct 10, 2015

BlueGas said:
Also what would the domain and range be for y = |x - 1|?

If there was also another question asking, for what values is the f(x) function increasing? What would the answer be to that?

$For the function $f(x) = |x-1|$? That is increasing on the interval $[1,\infty)$ (i.e. $x\geq 1$).$

InteGrand · Oct 10, 2015

rand_althor said:
If you draw a graph you should easily be able to figure out both the domain and range. If you don't have one here you go:

Domain: what possible x-values can be subbed in to the function?
Range: what possible y-values are outputted once these x-values are subbed in?

Now to know where f(x) is increasing, check a graph of f'(x) (you posted one in your original post). The function is increasing when f'(x)>0, and decreasing when f'(x)<0. For what x-values do these statements hold true in the graph of f'(x)?

$If you have the graph of $y=f(x)=|x-1|$, you don't need to bother checking $f^\prime$, since it's clear from the graph of $y=f(x)$ already that it's increasing on $[1,\infty)$.$

rand_althor · Oct 10, 2015

InteGrand said:
$For the function $f(x) = |x-1|$? That is increasing on the interval $[1,\infty)$ (i.e. $x\geq 1$).$

Why x greater than or equal to 1 and not just x greater than 1? The f'(x) graph BlueGas posted has a hole at x=1.

InteGrand · Oct 10, 2015

rand_althor said:
Why x greater than or equal to 1 and not just x greater than 1? The f'(x) graph BlueGas posted has a hole at x=1.

$Even if the derivative there were 0, we'd include it. The definition of a function $f$ being \textsl{increasing} on an interval $I \subseteq \mathbb{R}$ is that for all $x_1,x_2 \in I$, if $x_2\geq x_1$, then $f(x_2) \geq f(x_1)$ (for a function that is both continuous and has a continuous first derivative, this is equivalent to requiring $f^\prime (x) \geq 0$ for all $x\in I$; however, for our function here, the derivative is \textsl{not} continuous). This definition of ``increasing'' is satisfied on $[1,\infty)$, and this is the largest interval that it is satisfied in, so it is the answer to the question. If the function itself had a hole at $x=1$ (i.e. were undefined there), we would have omitted 1.$

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