obsidian phoenix
I need help to understand the answer key to this functions and graphs question. I do not understand why c is negative. I set f(x) = k and solve for k to get intersections, I think there should always be an intersection (and hence a solution for k) because the question says f(x) has a range of all real numbers. Hence I think the discriminant for k should be positive or 0.
boreal boneBOT
I need help to understand the answer key to this functions and graphs question. ...
+close
long zinc
I need help to understand the answer key to this functions and graphs question. ...
Sorry the text is rather small, and the picture gets blurry upon zooming
Could you post a zoomed in version? got a clear view now
violet meteorBOT
ℝafain
long zinc
This last part isn't valid, the preceding inequality is just true for any real number k
obsidian phoenix
So the first discriminant in terms of k is always positive, hence there should be no or 1 solution for k. Hence the second discriminant in terms of c should be negative (first discriminant is always positive and does not touch k axis) or 0 (1 solution for k, first discriminant touches k axis once). Right?
long zinc
Precisely
obsidian phoenix
I think I follow your approach but I do not understand what is incorrect in my reasoning in red in this picture. Could you help me identify the flaw in the red reasoning?
long zinc
but k in R is always defined
What do you mean by this?
obsidian phoenix
I initially set y = k and find intersections with y = f(x). I think there are always intersections and hence solutions for k because f(x) has a range of all real numbers.
obsidian phoenix
So the first discriminant in terms of k is always positive, hence there should b...
This appears to contradict the conclusion of the first approach.
c(c+2) cannot both be positive and negative?
long zinc
I think we are mixing up the graphs {(x, f(x)): x in R} and {(x, k): x in R}, where x is the independent variable and k is one (fixed) real number
and {k, (9+4c)k^2 + (6+4c)k + 1: k in R}, where k is the independent variable and varies over R
Depending on the value of c, this second parabola can very well not intersect the k-axis or just touch it at 1 point
long zinc
This appears to contradict the conclusion of the first approach.
Which is the first approach?
obsidian phoenix
First approach: The first discriminant in terms of k is always positive, hence there should be no or 1 solution for k. Hence the second discriminant in terms of c should be negative (first discriminant is always positive and does not touch k axis) or 0 (1 solution for k, first discriminant touches k axis once).
Second approach: I initially set y = k and find intersections with y = f(x). I think there are always intersections and hence solutions for k because f(x) has a range of all real numbers.
Conclusion of second approach: solutions for k always exist and hence discriminant in terms of k is positive or 0 so c(c+2) >= 0.
Problem: conclusion of approach 1 contradicts conclusion of approach 2. c(c+2) cannot both be positive and negative?
long zinc
Conclusion of second approach: solutions for k always exist and hence discrimina...
I do believe you are mixing up solution sets of different equations and/or inequalities
obsidian phoenix
obsidian phoenix
Question: I agree with the reasoning in approach 1. But I do not see the error in approach 2. What is wrong in the reasoning in approach 2 in the picture?
long zinc
The second approach is flawed, it conflates the existence of real solutions for x to the equation f(x) = k for every real number k,
to the existence of real solutions for k to the inequality (9+4c)k^2 + (6+4c)k + 1 >= = 0 for some particular c
Different conditions imply different feasibilities, simple as that
obsidian phoenix
Approach 2 says if there is always a solution for x for intersections between y = k and y = f(x), then there must always be solutions for y and hence k as y = k. This appears equally valid, why is it incorrect?
long zinc
When you say f(x) = k has a solution for every real number k, you are fixing k to consider the equation, and then saying there is x that satisfies that equation
Just that whichever k you fix on, the corresponding equation always has a solution for x
The aforementioned equation is equivalent to kx^2 - (3k+1)x - c(k+1) = 0 whenever x^2 - 3x - c is nonzero
Then when we consider the inequality (discriminant of this quadratic equation in x) >= 0, we are no longer only looking at one single k
Every real number k satisfies this inequality
Then we can plot the graph {(k, (9+4c)k^2 + (6+4c)k + 1): k in R}, whose second coordinate is always nonnegative
This being a parabola, it either touches the k-axis at one point or is always above the k-axis
{(k, (9+4c)k^2 + (6+4c)k + 1): k in R}
The graph takes any real input k (the independent variable); nonetheless (9+4c)k^2 + (6+4c)k + 1 = 0 may have no real solution or 1 real solution for k
long zinc
The second approach is flawed, it conflates the existence of real solutions for ...
Previously I messed up the equal sign here
So @obsidian phoenix , different equations, different variables to solve for, different solution sets
obsidian phoenix
Yes I follow that the 2 approaches lead to contradicting conclusions
But what is wrong with this reasoning?
long zinc
No... I am saying that approach 2 does not hold
When you say f(x) = k has a solution for every real number k, you are fixing k to consider the equation, and then saying there is x that satisfies that equation
The very first part does not solve for k, it fixes k and then solves for x
long zinc
> {(k, (9+4c)k^2 + (6+4c)k + 1): k in R}
The graph takes any real input k (the i...
Also here
A simpler analogous example would be, the function f(x) = x^2 + 1 takes in every real number x, but x^2 + 1 = 0 has no real solution for x
So "k in R is always defined" just means that quadratic function in k can take any real number in place of k, but does not guarantee the quadratic function has real roots
Imma try something extreme: you say "k in R is always defined", and think it implies that a quadratic function in k has roots
Then is every real number a root to that quadratic function? Wouldn't that just make it the zero function?
long zinc
Thank goodness I'm almost at wit's end 😭