Science:Math Exam Resources/Courses/MATH110/April 2013/Question 03 (d)

MATH110 April 2013

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Question 03 (d)
Let $f(x)={\frac {\sin(x)}{e^{x}}}$ on the interval $[0,\pi ]$. In this question you will sketch this function. (d) Determine where ƒ is concave up and where it is concave down. Does ƒ have any inflection points?

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Similar to the previous question, now find the second derivative, its possible inflection points, and test the sign of the second derivative between possible inflection points.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We know from the previous part of the question that $f'(x)={\frac {\cos x-\sin x}{e^{x}}}$ We take the derivative of $\displaystyle f'(x)$ to find: ${\begin{aligned}f''(x)&={\frac {(-\sin x-\cos x)e^{x}-(e^{x})(\cos x-\sin x)}{(e^{x})^{2}}}&={\frac {e^{x}(-\sin x-\cos x-\cos x+\sin x)}{(e^{x})^{2}}}&={\frac {-2\cos x}{e^{x}}}\end{aligned}}$ Setting $\displaystyle f''(x)$ equal to zero to solve for possible inflection points gives: $0={\frac {-2\cos x}{e^{x}}}$ or, equivalently, $\displaystyle \cos x=0$ Using the unit circle we see that x = π/2 is the only solution in $[0,\pi ]$ of the equation $\cos(x)=0$. This is our only possible point of inflection, because $\displaystyle f''(x)$ is continuous everywhere. Testing the values of $f''\left({\frac {\pi }{2}}\right)$ with the second derivative test gives: $[0,\pi /2)$ $x=\pi /2$ $(\pi /2,\pi ]$ $\displaystyle f''(x)$ - 0 + $\displaystyle f(x)$ concave down inflection point concave up Hence $\displaystyle f(x)$ has an inflection point at $x=\pi /2$.

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