a) Was there ever a time when the first function = the second function?
Was there ever a time when 16x/x^2+2 = 12x/x^2+1 ?
16x / x2 + 2 = 12x / x2 + 1 I am guessing that this is supposed to be...
16x / (x2 + 2) = 12x / (x2 + 1) Cross-multiply.
16x(x2 + 1) = 12x(x2 + 2) Distribute.
16x3 + 16x = 12x3 + 24x Subtract 12x3 from both sides.
4x3 + 16x = 24x Subtract 24x from both sides.
4x3 - 8x = 0 Factor 4x out of both terms.
4x(x2 - 2) = 0 Set each factor equal to zero and solve for x .
4x = 0 or x2 - 2 = 0
x = 0 or x2 = 2
x = ± √2
Yes, they are the same when x = 0 , x = √2 , and x = -√2 .
b) For what values of x is 16x/x^2+2 greater than 12x/x^2+1 ?
16x / (x2 + 2) > 12x / (x2 + 1)
Look at the graph here.
We can see 16x / (x2 + 2) is greater than 12x / (x2 + 1) for x > √2 and for -√2 < x < 0 .
You might just need to say positive possibilities: x > √2
I believe I have gotten the answer to this riddle! Using the slope formula, one can find the value for a, if a solution exists.
\(m=\frac{y_2-y_1}{x_2-x_1}\)
Just plug in the appropriate values for the x- and y-corrdinates and solve.
\(\frac{5a^2-3a^2}{3a+4-(2a+4)}=a+3\) | Now, solve for a. First, distribute the negation to all the terms inside the parentheses. |
\(\frac{5a^2-3a^2}{3a+4-2a-4}=a+3\) | Simplify the numerator and denominator in the fraction because both happen to have like terms. |
\(\frac{2a^2}{a}=a+3\) | Both the numerator and denominator can cancel out an "a." |
\(2a=a+3\) | Subtract a from both sides. |
\(a=3\) | Wow! There is actually a solution! |
In case you are wondering, this is correct. Click here to view the corresponding graph.