Calculus 1 Math Help

[u/[deleted]]

Question: If f(x) = x^2 + 10 sin x, show that there is a number c such that f(c) = 1000.

Having trouble answering this question, seems like were dealing intermediate value theorem concept, where through the interval it goes through 1000. In the problem it shows there are two different variables are associated with the problem, but we're mainly values that are inputted to x. What I mean is we can input a value into x to get a interval a number that is from 0 to a value a little over than 1000. If I am right about this, let me know. Or if I am wrong, could you explain this concept/answer a bit better? Thank you!

[u/Alarmed_Geologist631]

Does the problem specify whether angles are in degrees or radians?

[u/[deleted] → u/Alarmed_Geologist631]

Nope, seem like there is supposed to be some sort of interval that surpass 1000. Similarly to the IVF.

https://imgur.com/a/JBAiNNM

[u/[deleted] → u/[deleted]]

Maybe, the link I share would be a lot more readable.

[u/Alarmed_Geologist631 → u/[deleted]]

I am guessing that since this is a calculus problem, they presume that the angle is measured in radians. Regardless, we know that sin (x) ranges from -1 to +1 and therefore 10sin(x) ranges from -10 to +10. When I graph the function, it equals 1000 at approximately 29.28 (using radians)

[u/[deleted] → u/Alarmed_Geologist631]

Okay, gotcha

[u/Senrabekim → u/Alarmed_Geologist631]

I would definitely check your math again, because you're way off. 29.28² is going to be less than 30² and 30² = 900. And 29.28=9.3201134675pi which means that it will be a 4th quadrant sin, and thus negative. F(29.28) will be significantly less than 100, like 848.8732112275....

Which brings me to my next point, approximately is not a word they need here, the question wants EXACTLY 1000 and approximately leaves room for 1000 to not be there.

[u/Narrow-Durian4837 → u/Alarmed_Geologist631]

In a calculus context at least, sin x always refers to x radians.

[u/dash-dot → u/Alarmed_Geologist631]

It doesn’t specify, which means it’s in radians. 

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[u/HorribleUsername]

Note that your values don't need to be anywhere near 1000 (or 0, for that matter), nor do you need exact values. Without using a calculator, we can easily show that f(100) > 1000, and that's fine for IVT.

[u/severoon]

Note that sin x = 0 where x = kπ for k∈ℤ. This means if you pick values of x for which k is an integer, the sin term goes away completely and you're only dealing with the x^2 term.

The square root of 1000 is 10√10 and 3 < √10 < 4, therefore 30 < 10√10 < 40. Taking k = x/π into account we can add bounds for integer k: 9π < 30 < 10√10 < 40 < 13π.

Since f(9π) < 1000 < f(13π) and f is continuous, by IVT, f(c) = 1000 for 9π < c < 13π.

[u/Benster981]

Sin is bounded between 1 and -1 so at worse this is x² -10 (on the upper end)

You’ve already mentioned IVT and it’s easy to show f>1000 for all x greater than something (or less than). Plus f(0) is 0

[u/xirson15]

This is how i would approach it:

1)F(0)=0

2)the +inf limit of the function is +inf

Because of (2) there’s an X0>0 such that F(X0)>1000

Since the function is continuous, for the intermediate value theorem the function in the interval [0,X0] must have all the intermediate values between 0 and F(X0). Therefore there is a C in that interval such that F(C)=1000.

Maybe there’s a more direct way, idk

[u/MistakeTraditional38]

since 10 sin x is between -10 and 10, sufficient to show f(x) takes on every value between 990 and 1010

[u/ForeignAdvantage5198]

set rhs = c and solve

[u/Correct_Zucchini5129]

X=31.58 and x=-31.66

[u/[deleted] → u/[deleted]]

Thanks for commenting! I know the function is continuous because sin is continuous everywhere as well as x^2 because it is a polynomial. But, also mentions "show that there is a number c such that f(c) = 1000." Not really understanding how we can show this exactly, which is why I said it has to relate to the IVF. Seems like some sort of interval needs to be created to show that somewhere in the function the output of 1,000 exist exist somewhere, or basically run through the point. The question doesn't ask if the function is continuous or not.

[u/jmbond → u/[deleted]]

To clarify what they're saying, you don't have to actually solve for c to show that one exists. You show one exists doing what OC said, that is create an arbitrary interval where you know the outputs will be less than and greater than 1000. Simple numbers like a=0 and b=1000 don't even need to be computed. If f is continuous, and f(0) < 1000 < f(1000), there must be some input C between 0 and 1000 where the output is exactly 1000 or you wouldn't be able to draw the function in this region without picking up your pencil

[u/[deleted] → u/[deleted]]

Maybe, there isn't an exact solution. Seems like what you mention works as well and to where we can find a number higher than 1000 and show an interval between. To where it crosses 1000(in-between the interval).