A cross-section perpendicular to the rotation axis is a disk of radius $\sin x$. Disc integration therefore givesThe result is in cubic coordinate units; the full rotation is already accounted for by the disk area.

Use completing the square after taking out the leading coefficient:Thus $a=3$, $b=-2$ and $c=-1$. Since a real square is nonnegative, the global minimum is $-1$, attained at $x=2$.

The derivative is $dy/dx=3x^2$, giving slope twelve at $x=2$. The point on the curve is $(2,9)$. The point-slope equation of the tangent line is consequently

Write $z=x+iy$. The complex modulus condition is equivalent to $(x-1{)}^2+y^2=4$. Thus the locus is a circle of radius two centred at $(1,0)$ in the complex plane. Its extreme points are $(-1,0)$, $(3,0)$, and $(1,\pm 2)$.

This is a separable differential equation. Multiply by $e^{-y}$ and integrate:The initial condition gives $C=-1$. HenceFor a real solution the outer logarithm requires a positive argument. The maximal interval containing the initial point is therefore $\boxed{|x|<\sqrt{e^2-1}}$. At either endpoint $y$ diverges, so the real solution cannot continue through it with a finite value.

For $f(x)=\sqrt{x}$, the first two derivatives at one are $f^{\prime }(1)=1/2$ and $f^{\prime \prime }(1)=-1/4$. The Taylor expansion beginsAll three displayed terms are nonzero terms of the expansion about one.

Rationalize with the complex conjugate $4-3i$:Thus the real part and imaginary part are

Direct matrix multiplication givesSubtracting verifies $\boxed{A^2-4A-5I=0}$. This is also the Cayley-Hamilton theorem for the characteristic polynomial ${\lambda }^2-4\lambda -5$.

The quotient rule simplifies the derivative toIts zeros are $-5$ and $-1$, both in the domain. Substitution gives the two stationary pointsThe first is a local maximum and the second a local minimum, since the numerator of $y^{\prime }$ changes sign from positive to negative at $-5$ and from negative to positive at $-1$. The denominator of the original rational function vanishes atThe numerator is nonzero at both values, so these discontinuities are poles, not removable holes.

The numerator is the derivative of the denominator. Using the chain rule in reverse gives the antiderivativeThe absolute value makes this real on every interval where the denominator is nonzero. The integration constant may be chosen independently on the intervals separated by $x=(-3\pm \sqrt{17})/2$.