SOLUTION: Suppose that the function f is defined for all real numbers as follows. f(x)= -x^(2)+10 if -4<=x<4 -5-x if x>=4 Graph the function f. Then determine whether o

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Question 1165235: Suppose that the function f is defined for all real numbers as follows.
f(x)= -x^(2)+10 if -4<=x<4
-5-x if x>=4
Graph the function f. Then determine whether or not the function is continuous.

Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52855)   (Show Source): You can put this solution on YOUR website!
.
Suppose that the function f is defined for all real numbers as follows.
f(x)= -x^(2)+10 if -4<=x<4
-5-x if x>=4
Graph the function f. Then determine whether or not the function is continuous.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

In each of the two domains, the function is defined as a polynomial.


Therefore, inside each domain the function is continuous (since a polynomial is 
a continuous function, as you should know from Calculus).


Hence, to be a continuous function over the entire domain, the necessary and sufficient condition 
for the given function is to have the limit at x= 4 from the left to be equal to its value at x = 4.


The limit at x = 4 from the left side is  -x^2 + 10 at x ---> 4,  i.e. -4^2+10 = -16+10 = -6.


The value of the function at  x = 4  is  -5 - 4 = -9.


-9 =/= -6, so we conclude that the given function is not continuous at x = 4.

At this point, we complete our reasoning and answered the question.


//////////////////////////////////////


Notice that the problem's formulation has a deficiency.

Indeed, it says that "the function f is defined for all real numbers as follows . . . ",
but in reality it is defined only on the union of two intervals [-4,4) and [4,).

This union is the whole interval [,), but it is not the set of all real numbers,
as the problem proclaims. Thus, the problems wording is not perfect and is not accurate.


By such signs, every professional Math reader always can unmistakably recognize
"problems" that were written by an unprofessional "in a garage on the knee."


By the way, the first three words in the problem "Suppose that the" are excessive.
They are not necessary and can be omitted. Then the problem's appearance will be better.

It is a rule of Math writing: remove everything that does not matter.

In Math, following this rule is expressing of respect to a reader.



Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

This is the graph of the piecewise function.

The red parabola is due to the piece y = -x^2+10, when graphed on the interval -4 <= x < 4.
Graph y = -x^2+10 as normal, then erase the portions when x < -4 or x > 4.
The vertex is at (0,10).
Two other points on this parabola are (-1,9) and (1,9).

Two graphing tools I recommend are Desmos and GeoGebra
Or you can use something like a TI83.
Another alternative is to use graph paper to do everything by hand.

The blue line is the piece y = -5-x on the interval x >= 4.
Two points on this blue line are (4,-9) and (5,-10).

There is a red closed endpoint at (-4,-6)
There is an open hole at (4,-6) marked in red.
There is a blue closed filled in endpoint at (4,-9)
Another way of saying "closed endpoint" could be "filled in endpoint".

The open endpoint is due to the lack of "or equal to" in the inequality sign.
This is why we exclude x = 4 from the red parabola.
In contrast every other endpoint is included because there is an "or equal to".

We can clearly see the graph is not continuous.
There's a jump at x = 4 when going from the red curve to the blue line.

--------------------------------------------------------------------------

To determine this algebraically, without needing a graph, plug x = 4 into both pieces to see what results.
I'll label the pieces g(x) and h(x).
g(x) = -x^2+10
g(4) = -4^2+10
g(4) = -16+10
g(4) = -6
This leads to the location (4,-6) which is the red open hole on the graph.

And,
h(x) = -5-x
h(4) = -5-4
h(4) = -9
This leads to the blue filled in endpoint at (4,-9)
The results -6 and -9 do not agree on the same number, so this is a non-graph confirmation that the piecewise function is not continuous.
To be continuous, we would need g(4) = h(4) to be the case.


Similar practice problems are at the following links
https://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.1206484.html
https://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.1188881.html
https://www.algebra.com/algebra/homework/Rational-functions/Rational-functions.faq.question.1200349.html

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