Calculus (Maths Methods Units 3 and 4) quiz

$3e^{3x} \sin(x)$

$3e^{3x} \cos(x) + e^{3x} \sin(x)$

$3e^{3x} \sin(x) + e^{3x} \cos(x)$

$e^{3x} (3 \sin(x) - \cos(x))$

$\frac{1}{x^2 + 1}$

$2x \ln(x^2 + 1)$

$\frac{2x}{(x^2 + 1)^2}$

$\frac{2x}{x^2 + 1}$

$\frac{1}{2} \sin(2x) + C$

$\sin(2x) + C$

$-\sin(2x) + C$

$-\frac{1}{2} \sin(2x) + C$

$\frac{t^3}{3} + 4t$

$\frac{t^3}{3} - 4t$

$2t$

$\frac{t^2}{2} - 4$

$4(3x^2 + 1)^3$

$(6x)(3x^2 + 1)^3$

$24x(3x^2 + 1)^3$

$12x(3x^2 + 1)^4$

$0$

$2$

$\frac{8}{3}$

$3$

A local maximum

A local minimum

A point of inflection

An asymptote

$\frac{1}{12}$

$\frac{1}{6}$

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{1}{\cos^2(2x)}$

$-\frac{2}{\cos^2(2x)}$

$\frac{2}{\cos^2(2x)}$

$2\tan(2x)$

$x = 0$

$x = e$

Infinity

$x = 1$

$f(x) = 2x^3 - 4x + 5$

$f(x) = 2x^3 - 4x + 3$

$f(x) = 2x^3 - 4x$

$f(x) = 3x^3 - 4x + 4$

I will find the maximum by substituting several whole-number values of $x$ into the expression and keeping whichever one happens to give the biggest answer, since the largest output I test must be the true maximum.

Let $x$ denote the variable and let $V(x)$ be the volume. Then $V'(x) = 0$ identifies stationary points, and the sign of $V''(x)$ confirms a maximum within the feasible interval.

Optimisation just requires algebra, so I will expand and rearrange the expression as far as possible and then read off the largest term as the maximum value.

First draw a diagram, then guess the dimensions that look about right and check that the shape seems sensible before writing down the final answer.