Finding inflection points- where did I miscalculate?

Question

$3 \sin x- \sin^3x$ on $[0, 2\pi]$

First derivative is :$3 \cos x-3 \sin^2x \cos x$.

I have written it like this: $3 \cos x(1-\sin^2x)$.

Using the identity $1 - \sin^2x = \cos^2x$, I get: $3 \cos x \cos^2x$.

I mark '$\cos x$' as 't' and then: second derivative is $9t^2$.

Comparing it to $0$, I get $x=\pi/2$ and $3\pi/2$.

In the workbook, it didn't assign value $t$ as $\cos x$ but instead went like this:

$-9 \cos^2x\sin x$

That is the second derivative and the span of solution it gets is: $x=0,x=\pi/2,x=\pi$ and $x=3\pi/2$. I fully understand this way, but why does marking '$\cos x$' as '$t$' get me a different span of solutions than in the workbook?

Answer 1

After you apply the clain rule to obtain the second derivative (see N. F. Taussig's description below), your derivative should have been $f''(t) = 9t^2 t'$. Then, you can split the resulting equation $-9 \cos^2 x \sin x = 0$ into two parts to find the inflection points.

For your first part ($-9 \cos^2 x = 0$) the two solutions you obtained are absolutely correct ($\pi/2 \text { and } 3\pi/2$). For the second part ($\sin x = 0$) we have $x = 0 \text { or } x = \pi$. Plugging in $x = 0$ gives us no change in sign, so no inflection point, but plugging in $\pi$ does; hence the three inflection points are $\pi/2, \pi \text {, and } 3\pi/2.$

Answer 2

You failed to apply the Chain Rule.

Chain Rule. Let $h = g \circ f$. Suppose that both the derivatives $f'(x)$ and $g'(y)$ exist, where $y = f(x)$. Then the derivative $h'(x)$ also exists and is given by the formula $h'(x) = f'(x) \cdot g'(f(x))$.

You differentiated $y = 3\sin x - \sin^3x$ to obtain \begin{align*} y' & = 3\cos x - 3\sin^2x\cos x\\ & = 3\cos x(1 - \sin^2x)\\ & = 3\cos x(\cos^2x)\\ & = 3\cos^3x \end{align*} Thus, the second derivative is \begin{align*} y'' & = 9\cos^2x(-\sin x)\\ & = -9\cos^2x\sin x \end{align*} You made the substitution $t = \cos x$ to write $y'(t) = 3t^3$. However, observe that since $t = \cos x$, what we really have is $y'(t(x)) = 3[t(x)]^3$. Applying the Chain Rule yields \begin{align*} y''(t(x)) & = 9[t(x)]^2[t'(x)]\\ & = 9\cos^2x(-\sin x)\\ & = -9\cos^2x\sin x \end{align*} in agreement with the result we obtained above.