+2446 #10, why not?
According to the original relation, \(f(x)=\sqrt[3]{x-4}\). There are a few steps that need to be done in order to find the inverse.
#1) Convert to y=-notation. I find that it is easier when working with this notation. This is quite a simple step, wouldn't you agree?
\(f(x)=\sqrt[3]{x-4}\Rightarrow y=\sqrt[3]{x-4}\)
#2) Now, replace all instances of a "y" with an "x," and replace all instances of an "x" with a "y." This is not a difficult step, either, as you might imagine.
\(y=\sqrt[3]{x-4}\Rightarrow x=\sqrt[3]{y-4}\)
#3) Solve for y. This step can range in difficulty. In this case, it is quite a simple step.
| \(x=\sqrt[3]{y-4}\) | Cube both sides. |
| \(x^3=y-4\) | Add 4 to both sides to isolate y. |
| \(y=x^3+4\) | |
#4) Convert back to function notation since the original problem was given in function notation. This is quite simple, too.
\(y=x^3+4\Rightarrow f^{-1}(x)=x^3+4\)
#5) Correspond your answer with the answer choices given!
+2446 To solve for a variable, in this case, simpl means to isolate it. That is the goal of solving for a variable.
1)
| \(C=K\left(\frac{Rr}{R-r}\right)\) | Let's first multiply both sides by (R-r). This will eliminate the denominator, which will ease the process of solving for "K." |
| \(C(R-r)=KRr\) | Divide by Rr to isolate K. |
| \(K=\frac{C(R-r)}{Rr}\\\) | You could distribute, if you so desired, but it is not necessary. This just creates more work for yourself, anyway. This is the final answer. |
2) This one will be harder since two "R's" exist in the original equation. This generally signals for the use of grouping.
| \(C=K\left(\frac{Rr}{R-r}\right)\) | Let's do the exact same thing as before; multiply by R-r. |
| \(C(R-r)=KRr\) | Let's distribute the C into the binomial. This will allow us to have two terms with "R's" |
| \(CR-Cr=KRr\) | Let's subtract KRr from both sides. Let's add Cr to both means meanwhile. |
| \(CR-KRr=Cr\) | Using grouping, we can change this into one "R." |
| \(R(C-Kr)=Cr\) | Now, divide by C-Kr to isolate "R." |
| \(R=\frac{Cr}{C-Kr}\) | This is finished because we have isolated the R. |
3) We will have to utilize the same technique in number 2 in order to isolate "f."
| \(F=\frac{fg}{f+g-d}\) | Let's multiply the denominator on both sides. This will get rid of any and all pesky denominators in this equation, |
| \(F(f+g-d)=fg\) | There is a "g" in the trinomial, so we will have to expand that part. |
| \(Ff+Fg-Fd=fg\) | Move all terms with a factor of f on the left hand side. Any other terms are moved to the right hand side. |
| \(Ff-fg=-Fg+Fd\) | Factor out an "f" from both terms from the left hand side. |
| \(f(F-g)=-Fg+Fd\) | Finally, divide by F-g to isolate the wanted variable completey. |
| \(f=\frac{-Fg+Fd}{F-g}\) | No more simplification is possible here. Leave the fraction as is. |
+2446 For the second question, it is possible to utilize something known as the Pythagorean Inequality Theorem. The theorem actually has two parts, but the relevant portion for this particular problem is that if the square of a triangle's longest side length is greater than the the sum of the squares of the shorter side lengths, then the triangle is obtuse. This wording is quite verbose, I know, so I decided to create a visual aide as a reference for you!
Let's first try and identify what the longest side length is of this particular triangle is. We know that 16 and 21 are already defined lengths, but we also have the unknown side length, which is labeled as s. I will follow the visual aide above, and I will consider the longest side length to be of length "c." In the visual aide above, the shorter side lengths are labeled "a" and "b."
It is easy to exclude the side with length 16 as the longest side length since we know that there exists a side with length 21, which is longer. However, beyond this, there are two possibilities. The possibility is that:
Of course, we do not know which one is the longest side, so we will just have to assume both cases. Let's begin with assuming that "s" is the longest side, or our "c."
| \(c^2>a^2+b^2\) | "s" will be the longest side, so plug it into "c." Plug the other two side lengths in any order. | ||
| \(s^2>21^2+16^2\) | Now, solve for "s" to see what the possibilities are. | ||
| \(s^2>441+256\) | |||
| \(s^2>697\) | Take the square root of both sides. This results in an absolute value inequality. | ||
| \(|s|>\sqrt{697}\) | Of course, the absolute value splits the equation into two parts. | ||
| Of course, let's remember that "s" is an integer side length of a triangle, so "s" must be positive. We can reject the second inequality because those values include ones less tha zero. | ||
| \(s>\sqrt{697}\) |
The above algebra only solved the first case where we assumed that "s" is the longest side. Now, let's assume that the side with length 21 is the indeed the longest.
| \(21^2>16^2+s^2\) | Let's solve for "s" here by simplifying first and foremost. | ||
| \(441>256+s^2\) | Subtract 256 from both sides. | ||
| \(185>s^2\\ s^2<185\) | For me, I can interpret the inequality better when the variable is placed on the left hand side of the equation. Take the square root of both sides. | ||
| \(|s|<\sqrt{185}\) | Now, solve the absolute value inequality. | ||
| Unlike the previous inequality, we can combine this into one compound inequality. | ||
| -√185 < s < √185 | Of course, yet again, "s" is a side length, so it should be greater than zero. | ||
| 0 < s < √185 |
Of course, let's take clean this up. We know, by the given information, that we only care about integer solutions, so let's calculate the radicals.We now have set restrictions for "s." \(\)
√697 ≈ 26.40
√185 ≈ 13.60
We can then make the restrictions the following
We are not done yet, though! We have to take into account the side length restriction that Cpill figured out in part a! He found that 5 < s < 37
Therefore, integer sides with lengths 6 to 13 or sides with lengths 27 to 36 are possible. We can find the total number of integer solutions by figuring out how many integers are in this range.
From [6,13], there are 8 integers
From [27,36], there are 10 integers
Altogether, there are 18 integer solutions.
*Unfortunately, LaTeX appears to be funky when compound inequalities are introduced, so I have resorted to a different method to portray them.
