f(x)=x^2exp(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.

Jaxson Mack

Jaxson Mack

Open question

2022-08-15

f ( x ) = x 2 exp ( sin ( x ) ) cos ( x ) on the interval [ 0 , π / 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.

Answer & Explanation

heiftarab

heiftarab

Beginner2022-08-16Added 10 answers

The function x 2 e x p ( s i n ( x ) ) is increasing, the function c o s ( 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 ( π / 2 ) 2 e. Therefore it has precisely one zero point.

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