Functions - Range and set of values

Question

Thank you for taking your valuable time to review my question. I am really stuck with the below questions, and below the question I have written what I have done so far (my chain of thought). If you would be able to tell me what I have done is right/wrong, and how the end answer is reached including working that'd be appreciated. (Please don't laugh at what I've done, I know some things may sound silly!, and sorry!)

1) The functions $f$ and $g$ are defined as follows: $f(x)= x^2-2x $ which is $x \in\mathbb R$ and also $g (x)= 2x + 3$ which is $x\in \mathbb R$

i) Find the set of values for which $f(x) > 15 $.

(I have NO CLUE how to approach this question, some pointers are appreciated). Is it that ANY $x$ value which fits in the equation must be larger than $15$? But wouldn't there be many sets of values?

ii) Find the range of $f$ and state with a reason whether $f$ has an inverse.

The minimum point is $(1, -1)$ therefore the range is $f(x) \ge -1$ correct? And yes it does have an inverse because the inverse is the other minimum point?

iii) Show that the equation $gf(x) = 0$ has no real solutions.

The square of no real number gives a negative integer?. so the equation has no real solutions?

Thank you so much guys! I've attempted to show my chain of thought as much as possible because I don't want it to seem that I'm getting you to 'do my homework' (which it isn't as it is actually just practise questions) :) and I have made an attempt as best as possible. With an exam coming up, I hope that doing these additional questions will help.

Thanks!

Answer 1

For part i)

Try setting $f(x)=x^2-2x=15$. Since inequalities are tough to deal with, it's usually better in my opinion to use equalities and then think afterwards what's going on on either side of your critical values (the values of x which make your equality, $x^2-2x=15$ hold). In this case you find: $x^2-2x=15\Rightarrow (x-5)(x+3)=0$ meaning $x=5$ or $x=-3$. Plug in values on either side of each of these numbers to figure out what's going on: $x=4\Rightarrow x^2-2x=16-8=8>0$ which then tells you that, replacing equality with the inequality you want, $f(x)>15$. I'll leave the other cases to you.

for part ii)

As others are mentioning in the comments, try graphing the function (think about what it looks like) but also noticing that you have an $x^2$ term could be helpful; this means you are probably losing data, depending on where your function is defined. If I tell you that $x^2=1$, and I asked you what x was, you wouldn't be able to tell me if it was 1 or -1. This tells you that the funciton doesn't have an inverse, there is no road back after you put an input through your function machine.

for part iii)

First plug things in and see what comes out: $g(f(x))=g(x^2-2x)=2(x^2-2x)+3=2x^2-4x+3$. Setting this equal to zero gives:

$2x^2-4x+3=0\Rightarrow \frac{1}{4}(4\pm\sqrt{4-24)}=x=\frac{1}{4}(4\pm\sqrt{-20)}$ by the quadratic formula. As you can see, you have a negative number in the discrimant, which means your answer for x is not real.

Hope this helps, and don't worry about looking stupid. You put effort in, that's what matters here.

Answer 2

$i)x^2-2x-15>0$
The left hand side has two roots $3$ and $-5$ that means the l.h.s. has zeroes at those two roots.Now if you calculate l.h.s. at $0$ you would get $-15$ which is $-$ive .If you imagine the curve of l.h.s. you can easily find that it would cut x axis two times and between $3$ and $-5$ it has to be $-ive$ and has to go to positive before $-5$ and after $3$
$ii)$$$f(x)=x^2-2x \implies f'(x)=2x-2=0 \implies x=1$$ If you double differentiate $f(x)$ and get $f''(x) $ at $x=1$ you get positive value which means at $1$ it is minimum i.e. $f(1)$ is minimum and as you can see that f(x) is directly proportional to $x$ so it is maximum at $\infty$ so we get the range of it ($f(\infty)$ is maximum and $f(1)$ is minimum). <br> The inverse of $f(x)$ doesn't exist as for one value of $y$ we get two values of $x$ which is against the definition of function. <br> $iii)$$g(f(x))=2(x^2-2x)+3=0$ ,Now solve as qbert's answer. That's it
But i didn't understand how did you attempt the third part and put the argument?

Answer 3

Let $f(x) = x^2 - 2x$ and $g(x) = 2x + 3$ be defined on $\mathbb{R}$.

Find the range of $f$.

If we complete the square on $f(x) = x^2 - 2x$ to transform the equation into vertex form, we obtain \begin{align*} f(x) & = x^2 - 2x\\ & = x^2 - 2x + 1 - 1\\ & = (x - 1)^2 - 1 \end{align*} from which we can see that $f$ has a minimum value of $-1$ at $x = 1$. Its range is $[1, \infty)$ since the function increases without bound as $|x| \to \infty$.

Find the set of values for which $f(x) > 15$.

To determine where $f(x) > 15$, we must solve the inequality $(x - 1)^2 - 15 > 1$. \begin{align*} (x - 1)^2 - 1 & > 15\\ (x - 1)^2 & > 16\\ |x - 1| & > 4 \end{align*} Solving the absolute value inequality yields $x - 1 > 4$ or $x - 1 < -4$. \begin{align*} x - 1 & > 4 & x - 1 & < -4\\ x & > 5 & x & < -3 \end{align*} Hence, the solution set to the inequality is $[5, \infty) \cup (-\infty, -3]$.

State with a reason whether $f$ has an inverse.

The function does not have an inverse function since for each value of $y > -1$, there are two values of $x$ such that $f(x) = y$. For instance, $f(-3) = f(5) = 15$, so we cannot uniquely define $f^{-1}(15)$. For an inverse function to exist, for each value of $f(x)$, there must be a unique value of $x$.

Show that the equation $(g \circ f)(x) = 0$ has no real solutions.

By definition, \begin{align*} (g \circ f)(x) & = g(f(x))\\ & = g(x^2 - 2x)\\ & = 2(x^2 - 2x) + 3 \end{align*} Completing the square yields \begin{align*} (g \circ f)(x) & = 2(x^2 - 2x + 1) - 2 + 3\\ & = 2(x - 1)^2 + 1 \end{align*} The function $g \circ f$ has range $[1, \infty)$ with its minimum value of $1$ occurring at $x = 1$. Since $0$ is not in the range, the equation $(g \circ f)(x) = 0$ has no real-valued solutions.

Alternatively, set $(g \circ f)(x) = 0$ and solve for $x$. We obtain \begin{align*} (g \circ f)(x) & = 0\\ 2(x - 1)^2 + 1 & = 0\\ 2(x - 1)^2 & = -1\\ (x - 1)^2 & = -\frac{1}{2} \end{align*} Since no real number has a negative square, $g \circ f$ has no real roots.