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I've stumbled a lil confusion across this problem ,

$$f(t)=t^3-t^2-10$$

Where I need to plot the points $-3\le t\le 3$

I've done so and I've got these values of the function :

f(-3)= -46
f(-2)= -22
f(-1)= -12
f(0)= -10
f(1)= -10
f(2)= -6
f(3)= 8

So by this I understand that the zero lies between t=2 and t=3 , so the interval must be (2,3) . The practice question explanation says .

How many real zeros does it have? What interval do the zeros lie in? The endpoints of the intervals should be rounded to one decimal place and the interval length must be one tenth. Separate your answers using semicolons.

Well It does go through zero so It must be one.

To accomplish the interval length to one tenth I followed this procedure.

Found a middle value between the first two initial intervals and used it on the function of t .

$$f(2.5)=-0.625$$

As the result was negative must be the new lower interval so now I had $(2.5,3)$

After it to short the interval distance even further I chose 2.7 to be the middle-man between the 2 values.

$$f(2.7)=1.783$$

It's positive so I thought this would be the new upper limit. So as far I've got $(2.5,2.7)$

And now the final middle man between the interval.

$$f(2.6)=0.816$$

It's a positive so it must be the upper interval.

So I finally got to $$(2.5,2.6)$$

After this, to try and input the result as the required form that the question said I submitted the previous interval followed by the interval of the results of the functions of 2.5 and 2.6 as

$$(2.5,2.6);(-0.6,0.8)$$

(rounded up the second interval, as the problem said)

But, as good as it seemed to me the grading script is throwing back at me

The zero doesn't lie within your interval. For the lower bound, in this particular case, the function should be negative, and for the upper it should be positive.

I believe I don't quite get what the endpoints are, am I right about my answer? Does it really lays between those two points, or have I done any mistake.

Any kind of help is super appreciated! Thank you!

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  • $\begingroup$ It looks to me like you're asking about the script rather than the math. You should be asking this to your professor/teacher/the site administrator/whomever manages the grading script. $\endgroup$ – davidlowryduda Feb 11 '14 at 17:00
  • $\begingroup$ @mixedmath Oh I'm sorry I missed the whole point, edited post. $\endgroup$ – Joel Hernandez Feb 11 '14 at 17:09
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"Well It does go through zero so It must be one." - Actually, so far this only shows at least one. In priciple, there might be three zeroes in $(2,3)$ without you noticing it by the values at $t=2$ and $t=3$. Or there might be two zeroes in $(1,2)$.

The way I read the problem description, the answer should simply be $$(2.5,2.6)$$ or (if the system rather expects a list of endpoints $$2.5;2.6 $$ Or, since the question(s) are in fact "How many real zeros does it have? What interval do the zeros lie in?" $$1;(2.5,2.6)$$

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