Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Have the following

$f(x)=x^2 \exp(\sin(x))-\cos(x)$ on the interval $[0,\pi/2]$, I have shown the function is continuous and that there is at least one solution on the interval via using IVT, I know I have to find another solution in the interval such that $f(x)<0$, $f(x)>0$ and then $f(x)<0$, just wondering what is the best way to approach, Im sure there must be a more effective way than just to number crunch, could we consider MVT on the interval, many thanks in advance.

share|improve this question
1  
By "one solution," presumably you mean one solution to $f(x)=0$? Pedantic, yes, but a function is not an equation. –  Thomas Andrews May 1 '12 at 15:29
4  
Take the derivative. You will get a positive impression. –  André Nicolas May 1 '12 at 15:30
    
Why do you think there is more than one solution to $f(x)=0$? –  Thomas Andrews May 1 '12 at 15:31
    
@AndréNicolas many thanks for the input. –  user24930 May 1 '12 at 15:42
    
The answer of @Simon Markett is much better than my comment, and I am not referring to my feeble pun. Sure, if you find the derivative, it is obvious that it is positive, end of story. But looking directly at the two functions is the "right" approach, derivative is mechanical. Look and then (if necessary) compute is much better than compute and then look. –  André Nicolas May 1 '12 at 15:48

1 Answer 1

up vote 3 down vote accepted

The function $x^2exp(sin(x))$ is increasing, the function $cos(x)$ is decreasing in your interval. Hence the combined functions is increasing on your interval. It starts with a value of $-1$ and goes up to $(\pi/2)^2e$. Therefore it has precisely one zero point.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.