Question

find the relative extrema of f(x)=-x^2+4x+5
i found the x intercepts, and the y intercepts
and the derivative
and the critical number
but I need to find the relative extrema, but i dont know how to or what it is exactly

Answer 1

Its where ever the derivative crosses the x axis

Answer 2

relative extrema are mins and maxes

Answer 3

ok, thanks. so how do i find it? find the x intercept of the derivative?

Answer 4

You find the zeros but you need to make sure it crosses the x-axis.

Answer 5

Like x^2 has a zero at x = zero but it doesnt cross the x axis

Answer 6

it is a parabola,
you certainly don't need calc for that

Answer 7

this one opens down
it has a maximum therefore
the maximum is the second coordinate of the vertex

Answer 8

the first coordinate of the vertex is always \(-\frac{b}{2a}\) and the second coordinate is what you get when you replace \(x\) by the first coordinate

Answer 9

okay
so
the extrema of this function is a maximum, because it is a parabola that opens down
to find it, i use -b/2a
im confused, what do you mean by first and second coordinate of the vertex? isnt it just one vertex?
and, if i use -b/2a, do i plug in the values from the derivative, or the original function?

Answer 10

Have you ever done "Completing the Square" method to solve an equation of a form
ax^2+bx+c=0 ?

Answer 11

yeah.
but i dont remember it well
you add the c to the other side
then
i forgot :p

Answer 12

I will show you an example.
And will go through the calculus-wise and the algebraic way of doing it.

Answer 13

\(\color{black}{\large \rm Example:}\) \(\tiny \\[0.5em]\)
Find the absolute minimum of \(\color{black}{\displaystyle f(x)=x^2+6x-4}\).
--------------------------------------------------

\(\color{black}{\large \rm Algebraic~~Solution:}\) (By Completing the square) \(\tiny \\[0.5em]\)
\(\color{black}{\displaystyle f(x)=x^2+6x-4}\)
\(\color{black}{\displaystyle f(x)=x^2+6x+\left(\frac{6}{2}\right)^2-\left(\frac{6}{2}\right)^2-4}\)
\(\color{black}{\displaystyle f(x)=x^2+6x+9-9-4}\)
\(\color{black}{\displaystyle f(x)=(x^2+6x+9)-13}\)
\(\color{black}{\displaystyle f(x)=(x+3)^2-13}\)
and you know that \(\color{black}{\displaystyle (x+3)^2\ge0}\), so the smallest value is \(-13\) and it occurs at \(x=-3\).

Answer 14

\(\color{black}{\large \rm Calculus~~Solution:}\) (by taking the derivative) \(\tiny \\[0.5em]\)
\(\color{black}{\displaystyle f(x)=x^2+6x-4}\)
\(\color{black}{\displaystyle f'(x)=2x+6}\) (applied power-rule, and derivative of constant is 0)
\(\color{black}{\displaystyle 0=2x+6}\) (setting the derivative =0)
\(\color{black}{\displaystyle x=-3}\).

This is the only critical number, and it results from \(f'(x)=0\). So, it's either a maximum or a minimum. How do we know which one is it?? Well, without being hypertechnical, just take any very close value and plug it into the function.

I will plug it into the function, but I will use my algebraic version where I completed the square - since this is easier.
\(\color{black}{\displaystyle f(-3.05)=((-3.05) + 3)^2-13}\)
\(\color{black}{\displaystyle f(-3.05)=-13+0.0025>-13=f(-3)}\)

So obviously since the point right near x=-3 is greater, you know that x=-3 is not a maximum, but a minimum of the function.

Answer 15

Sorry for the wording (if this was a face-to-face tutoring I would just explain this verbally)

Answer 16

So, you basically have the choice (for this particular case)

(1) Find the vertex of the parabola.
\(-\) If it opens down, it's the maximum.
\(-\) If it opens up, it's the minimum.

(2) Find the critical numbers, and then the maximum exactly the way you are currently learning to do.

Answer 17

Ok
Thanks
I have some questions
the completing the square part: i know you did (b/2)^2 but, i dont remember doing (b/2)^2 - (b/2)^2 when I learned it
so i dont know how you know "the smallest value is −13 and it occurs at x=−3."
ummm
oh i guess thats my only question
i mean, wait, i think its easier in this case to use the sign of the function to find out if its a max or a min
but your method can be used for when its not a parabola?

Answer 18

For question 1:

Have you ever heard the technique of multiplying times "magic 1" (something like times y/y)? Why do we multiply times top and bottom? Not to change the values and relationships in the equations. Correct?

Answer 19

So, here we add and subtract that (b/2)^2, so that we don't change the values and relationships between the variables in the equation. (The positive (b/2)^2 will go to help us make the perfect square trinomial, and the negative (b/2)^2 will contribute to the vertical shift.

Answer 20

In other words, we have two tricks in our sleeve.
(1) Multiplying times something on top and bottom (or times blank/blank).
(2) Adding a magic zero (something like adding R - R which doesn't affect our equation)

Answer 21

Tell me if you are good so far ... (as to, whether or not you understand why I add and subtract (b/2)^2) ?

Answer 22

Oh yes, I see.

Answer 23

So, this way we re-write the function as
\(f(x)=(x+3)^2-13\)

Answer 24

and then, it is relevantly simple to observe that your
function doesn't have a maximum, and the smallest point
is when \(x+3=0\) --> and \(y=-13\).

Answer 25

Any questions about this?
(We can prove anything provable as regards to this)

Answer 26

Oh, in your example, you can do it with either approach (using calculus or algebra), with an exception that you are asked for the maximum (because your function doesn't have a minimum, and one clue to this is that it is concave down for all x)

Answer 27

Uh, yeah, sorry I dont know why "it is relevantly simple to observe that your
function doesn't have a maximum, and the smallest point
is when x+3=0 --> and y=−13."
Sorry :p I just dont see how based on the solution from the completing the square thing, we concluded this.

Answer 28

The observation that \(f(x)=x^2+6x-4\) doesn't have a maximum,
is (basically) arising from \(\displaystyle \lim_{x\to~\pm\infty} f(x)=\infty\).

(and the domain is not restricted - it's all real numbers)

Answer 29

Now, what is the smallest point on \(f(x)=(x+3)^2-13\)?
Well, you know that the smaller the \((x+3)^2\) part is, the smaller is the function, and since the smallest possible value for \((x+3)^2\) is \((x+3)^2=0\) (which occurs at \(x=-3\)), therefore your minimum is at \(x=-3\).

Then, \(f(-3)=(\color{red}{-3}+3)^2-13=-13\), so our minimum point is \((-3,-13)\).

Answer 30

Ok I understand.

Answer 31

Very nice:)

Answer 32

So
Ill try to find the maximum now for my question?

Answer 33

Yes, you are finding the maximum for your question.

Answer 34

You will note also (by any function), that the relative maximum or relative minimum (or absolute min/max - provided it is not located at a closed boundary point), will be the point of horizontal tangent (Right? Just picture any function with many turns)

Answer 35

So, for this reason, we say that the \(x_{max}\) of the \(x_{min}\) is (possibly) a solution to \(f'(x)=0\).

Answer 36

I'm just making notes on concepts ...
(as it is now, I think we are pretty complete unless you say otherwise ... and if I'm online with some time, I will do my best to help you out)

Answer 37

what does xmax of the xmin mean?

Answer 38

(thanks for all your help btw, and yeah, if you can and want to. :) )

Answer 39

I am completing the square for my question right now
but i dont think (-x^2+4x+4) is factorable
well according to wolfram alpha it is (2+2 sqrt(2)-x) (-2+2 sqrt(2)+x)
but i would never have found that out on my own

Answer 40

\(x_{max}\) is the \(x\)-value that maximizes \(f(x)\).
i.e. A value of \(x\) for which the function \(f(x)\) is largest.

\(x_{min}\) is the \(x\)-value that minimizes \(f(x)\).
i.e. A value of \(x\) for which the function \(f(x)\) is smalles.

Answer 41

Now ... as regards to completing the square,.
First of all your function is: \(f(x)=-x^2+4x+5\) (+5 not +4)

Answer 42

What would you add to \(x^2-4x\) in order to complete the square?

Answer 43

i know
but
you have to do (b/2)^2
which makes it (-x^2+4x+4) -9
right?

Answer 44

You are implying the (b/2)^2 thinking correctly, but I feel there is a need for more clarification.

Answer 45

\(f(x)=-x^2+4x+5\)
\(f(x)=-(x^2-4x)+5\) (Note that: (b/2)\(^2\)=(-4/2)\(^2\)=4)
\(f(x)=-(x^2-4x\color{blue}{+4}\color{red}{-4})+5\)
\(f(x)=-(x^2-4x\color{blue}{+4})+(-1)(\color{red}{-4})+5\)
\(f(x)=-(\underbrace{x^2-4x+4}_{\small \rm perfect~sq.~trinomial})+4+5\)


\(\overbrace{\color{green}{\large f(x)=-(x-2)^2+9}}^{\color{purple}{ \large \rm (like~this)}}\)

Answer 46

Now you have to find the \(x\)-value for which \(f(x)\) is the largest.

Answer 47

(You can do this, by finding the \(x\) value that maximizes \(-(x-2)^2\). Then, as you find the \(x\), plug in the result to solve for the y-coordinate.)

Answer 48

why did you do + (-1)?

Answer 49

I'm just writing it clearly that I am taking (-4) out of those parenthesis

Answer 50

(still confused but)
okayyyy

Answer 51

Would the following be algebraically correct?
\(f(x)=\color{purple}{\bf B}(x^2-4x\color{blue}{+4}\color{red}{-4})+5\)
\(f(x)=\color{purple}{\bf B}(x^2-4x\color{blue}{+4})+(\color{purple}{\bf B})(\color{red}{-4})+5\)

Answer 52

And the same is here:
\(f(x)=\color{purple}{\bf -}(x^2-4x\color{blue}{+4}\color{red}{-4})+5\)
\(f(x)=\color{purple}{\bf (-1)}(x^2-4x\color{blue}{+4}\color{red}{-4})+5\)
\(f(x)=\color{purple}{\bf (-1)}(x^2-4x\color{blue}{+4})+(\color{purple}{\bf -1})(\color{red}{-4})+5\)

Answer 53

so the maximum y is 9

Answer 54

Yes, exactly!

Answer 55

and it occurs at what value of x?

Answer 56

(Then, write the point ... and this is the only exterma for f(x), because it is a quadratic equation, and thus will only have 1 turn.)

Answer 57

2
(2,9)

Answer 58

Yup, the point of extrema is (x,y)=(2,9) :)

Answer 59

Thank you verryyyyy much

Answer 60

You are welcome:) \(\color{black}{\displaystyle \psi\in[3,4]}\) :)

Answer 61

good luck with your math!