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How would I solve the following two questions.

Using the intermediate value theorem to show that there is a solution of the equation $\frac{sin^2x}{2}-x+1=0$ in the interval $[0,pi]$

I showed by the IVT there is a c in $[0,pi]$ give that c is zero because $0<1$ $0>-pi+1$ but I am not sure if it did this correctly.

2.My second question asked me to sketch a function that satisfies the following conditions or prove that it is impossible $f(x)$ is a continous function on $[0,2]$ its minimum value is $-3$ but it does not have a max value.

I said this is impossible because if $f(x)$ is continous on a bounded interval it must have both a max and min value.

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Your logic seems fine, though your write up for the first part needs more detail (Define the function $f(x)={\sin^2 x\over 2}-x+1$. Show the IVT, indeed, applies on the interval $[0,\pi]$. Then $f(0)>0$ and $f(\pi)<0$...) –  David Mitra Feb 2 '13 at 20:57

1 Answer 1

up vote 2 down vote accepted

Using the intermediate value theorem to show that there is a solution of the equation $\dfrac{\sin^2x}{2}-x+1=0$ in the interval $[0,\pi]$

"I showed by the IVT there is a c in $[0,\pi]$ give that c is zero because $0<1$, $0>-pi+1$ but I am not sure if it did this correctly."

Your process in answering this is just fine: just clarify the details:

Let $f(x)=\dfrac{\sin^2x}{2}−x+1.$
Show the Intermediate Value theorem does, in fact, apply on the given interval $[0,\pi]$: Then $f(0)>0$ and $f(\pi)<0$, using your computations...etc., including the details/justifications, you posted.

The second part looks just fine, as it is, as you explained exactly why it is not possible.

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I see thanks very much. –  Fernando Martinez Feb 2 '13 at 21:37
    
for my justifications I said because $f(0)=0<1$ and because $f(pi)=-pi+1<0$ –  Fernando Martinez Feb 2 '13 at 21:39
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Yes...that should be just fine! –  amWhy Feb 2 '13 at 21:42
    
@amWhy: Good to see you again while challenging the Maths. + –  Babak S. Feb 3 '13 at 15:23

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