Checking if the function has a zero in a specified interval.

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jkovie

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Hi I have a question which I have answered, just want to check if it's right and if I am thinking about it properly. Thanks.

Which of the following Functions has a zero in the interval [-1,3]?

a. F(x)= x/(x-4)
b. F(x)= (x^2) + (3X) + 2
c. F(x)=(x^2)/(x-2)
d. F(x)= cos(pi/x)
e. Both a and c do.

I answered a, c, and d

I plugged in -1 and 3 into each function and got these answers for each function:
a. f(-1) = 1/5 f(3) = -3
b. f(-1)=6 f(3)=5
c. f(-1)=-1/3 f(3)=9
d. f(-1) = -1 f(3)=1/2

So if f is continuous over all of x then a, c, and d should have zeros in their intervals.

Let me know if I am on the right track. Thanks.
 
jkovie said:
Hi I have a question which I have answered, just want to check if it's right and if I am thinking about it properly. Thanks.

Which of the following Functions has a zero in the interval [-1,3]?

a. F(x)= x/(x-4)
b. F(x)= (x^2) + (3X) + 2
c. F(x)=(x^2)/(x-2)
d. F(x)= cos(pi/x)
e. Both a and c do.

I answered a, c, and d

I plugged in -1 and 3 into each function and got these answers for each function:
a. f(-1) = 1/5 f(3) = -3
b. f(-1)=6 f(3)=5
c. f(-1)=-1/3 f(3)=9
d. f(-1) = -1 f(3)=1/2

So if f is continuous over all of x then a, c, and d should have zeros in their intervals.

Let me know if I am on the right track. Thanks.
Click to expand...

Your answer is correct - your reasoning is partially correct.

For b, you got two positive results and you concluded that the function does not have a root in the specified interval. But it is a parabola - so it can intersect the x-axis twice in the interval and still show the same sign at the end points. For example if the function was 2x2 - 3x + 1 then f(-1) = 6 and f(3) = 10. However, the function has a zero at x = 1/2 and x = 1.

So if the sign changes - it definitely has a zero (for continuous function). But if the sign does not change, it can have even number of zeros (0, 2, 4, etc.).
 
Which of the following Functions has a zero in the interval [-1,3]?

a. F(x)= x/(x-4)
b. F(x)= (x^2) + (3X) + 2
c. F(x)=(x^2)/(x-2)
d. F(x)= cos(pi/x)
e. Both a and c do.

So if F is continuous over all of x then a, c, and d should have zeros in their intervals.
Click to expand...

In questions a, c, and d, those are not continuous functions.
 
Last edited:
Thanks for the replies.

I should not have stated continuous. The question does not state that the functions are continuous or not. It just asks if there is a zero in the interval [-1,3]

So after looking at the graphs, all of them appear to have zeroes in the given interval. The question does not say only one zero or that the function needs to be continous so a,b,c, and d should be the answer.
 
Hello, jkovie!

Which of the following functions has a zero in the interval [-1,3]?

\(\displaystyle \begin{array}{cc}(a) & f(x)\:=\:\dfrac{x}{x-4} \\ (b) & f(x)\:=\: x^2 + 3x + 2 \\ (c) & f(x)\:=\:\dfrac{x^2}{x-2} \\ \\ (d) & f(x)\:=\: \cos(\frac{\pi}{x}) \\ \\ (e) & \text{Both (a) and (c) do.} \end{array}\)


I answered (a), (c), and (d).

I plugged in -1 and 3 into each function and got these answers for each function:

\(\displaystyle (a)\;f(\text{-}1) = \tfrac{1}{5},\;f(3) = \text{-}3\)

\(\displaystyle (b)\; f(\text{-}1)=6,\; f(3)=5\)

\(\displaystyle (c)\;f(\text{-}1)=\text{-}\frac{1}{3},\; f(3)=9\)

\(\displaystyle (d)\; f(\text{-}1) = \text{-}1,\; f(3)=\frac{1}{2}\)

So if \(\displaystyle f\) is continuous over all of \(\displaystyle x,\) . They are not all continuous.
then a, c, and d should have zeros in their intervals.

Let me know if I am on the right track. .Thanks.
Click to expand...

Testing endpoints is not a dependable procedure.

With \(\displaystyle (a)\;f(x) \,=\,\frac{x}{x-4}\), there is a zero at \(\displaystyle x = 0.\)
There is a vertical asymptote at \(\displaystyle x =4\), but it doesn't concern us.

With \(\displaystyle (b)\;f(x) \,=\,x^2+3x+2\), you made an error.
. . \(\displaystyle f(\text{-}1) = 0\)
There is a zero on the interval.

With \(\displaystyle (c)\;f(x) \,=\,\frac{x^2}{x-2}\), there is a zero at \(\displaystyle x = 0.\)
There is a vertical asymptote at \(\displaystyle x = 2;.\) the function is not continuous.
But it doesn't change our answer.

With \(\displaystyle (d)\;f(x) \,=\,\cos(\frac{\pi}{x})\), the function is undefined at \(\displaystyle x=0.\)
But we find that: \(\displaystyle f(2) \,=\,\cos(\frac{\pi}{2}) \,=\,0.\)
Hence, there is a zero at \(\displaystyle x = 2.\)
. . and there are more zeros at: .\(\displaystyle x \:=\: \pm\frac{2}{3},\:\pm\frac{2}{5},\:\pm\frac{2}{7} \text{ . . .}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Let's examine part (b) again.

Suppose we have: \(\displaystyle f(x) \:=\: x^2-x\)
Does it have any zeros on the interval [-1,3]?

By your method, you would get:
. . \(\displaystyle \begin{array}{ccc}f(\text{-}1) \:=\:2 \\ \;f(3) \:=\:6 \end{array}\) .and your answer would be No.

But \(\displaystyle f(x)\) has zeros at: .\(\displaystyle x \,=\,0,\,1\)
 
Thank you. So the way to approach this problem should be instead of plugging in the interval parameters, we should analyze the graph and pinpoint where the function has a zero and then check if it is within the interval parameters?
 
jkovie said:
I should not have stated continuous. The question does not state that the functions are continuous or not.
Click to expand...

Continuity should be taken into consideration nonetheless. Your reasoning is okay with linear functions, but functions with asymptotes need additional scrutiny. You need to be mindful of specific function behavior near asymptotes (going to positive or negative infinity from above or below, and that stuff); it will help you to not jump to false conclusion. These behaviors were covered in your precalculus course.

I suggested graphs for checking your answers because that's something that you may easily do on your own. You may not have access to graphs in an exam (or time to plot them all).

I see that soroban posted something that I was going to say. Word to the wise: Many calculus exercises are cooked up with shortcuts, for the savvy, so whenever you're puzzling over something within a domain that includes x = -1, 0, 1 sometimes it pays dividends to simply evaluate your function(s) at those three locations. Often, one can do this quickly in their head, by inspection.

Such is the case in this exercise, because F(0) is clearly a zero in questions a and c, and same for F(-1) in question b.

My suggestion may help you on an exam question some day, too, because more information up front is better when time counts.

You were confused by Pi/x radians, perhaps? Don't forget the trig basics, when working with trig functions in calculus!

Cheers :cool:
 
Hello again, jkovie!


To locate zeros of a function on an interval:

. . Solve \(\displaystyle f(x) \,=\,0.\)

. . If roots exist, select those
. . that fall within the interval.
 
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