How Does the Intermediate Value Theorem Confirm a Unique Zero in an Interval?

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Can someone get me started on this problem.

Show that the function has exactly one zero in the given interval.
f(x) = x^4 + 3x + 1 [-2, -1]

I know I have to use the Intermediate Value theorem in this but not really sure how to apply it. Also what does it mean when it says "a function has one zero" what is a zero?

Thanks in advance
 
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A zero is a number x such that f(x)=0. For example, the function f(x)=x^2 - 1 = (x-1)(x+1) has the zeros 1 and -1.

For the function above, it is hard to tell by some formula whether it has a real zero (not all functions have real zeros, and many have no zeros). But it is easy to find points where the function is negative and points where the function is positive. What does the intermediate value theorem tell you then?
 
donjt81 said:
Can someone get me started on this problem.

Show that the function has exactly one zero in the given interval.
f(x) = x^4 + 3x + 1 [-2, -1]

I know I have to use the Intermediate Value theorem in this but not really sure how to apply it. Also what does it mean when it says "a function has one zero" what is a zero?

Thanks in advance
What is f(-2)? What is f(-1)? What does that tell you (using the intermediate value theorem)?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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