How can I find all intervals containing solutions to the equation $4x^2-e^x=0$, without using a graph? I know how to solve this problem, but not without a graph. Thanks.
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I see that the first derivative yields $8x-e^x=0$. From here, by eyeballing it, without rigor of course, I can only tell that there are two critical points. And so at most three roots. Now how do I narrow down the roots? – Sep 29 '13 at 00:43
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Are you on a topic for root finding, like Bisection or Newton's? – Amzoti Sep 29 '13 at 00:47
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Yes, Bisection method. – Sep 29 '13 at 00:49
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Sorry, not really. I should be able to do this without bisection method though right? We are not required to use that here. – Sep 29 '13 at 00:52
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let us continue this discussion in chat – Sep 29 '13 at 00:52
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HINT:
Find the derivative of the function $f(x) = 4x^2 - e^x$ and set it to $0$ to find it's local minimum or maximum. You can solve the new equation using Lambert W function.
Then use the second derivative test to check if the point is maxima or minima. You should be able to obtain 1 local minima and 1 local maxima. First we have a minima, so that means that the function is strictrly decresing in that interval, now plug the minima into your equation to find if the value of the function is smaller than 0. Yes, it is and that means that there's one zero in that interval. Then check the interval between the minima and the maxima, if they are on different sides of the x-axis, the we have another zero. And at last check if the maxima is bigger than $0$, because in the interval $(maxima, \infty)$, the function is decreasing. So if for the maxima, function has value bigger than $0$, we have another zero in that interval.
So in the end there are 3 zeros of the function. And we have one in each of the intervals: $(-\infty,minima), (minima,maxima), (maxima,\infty)$
Stefan4024
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If you plug in values randomly for $f(x)$ and you're able to get $f(a).f(b)<0$ then $(a,b)$ is the interval in which roots of the equation would lie.
daOnlyBG
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Suraj sethi
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