Answer to Question #186703 in Calculus for Hetisani Sewela

Question

3. Let f be the function defined by

f (x) =( e^x/2)/x.

(a) Is there a y–intercept? (Explain)

(b) Determine the horizontal and vertical asymptotes (if any).

(c) Use the sign pattern for f^'(x) to determine

(i) the interval(s) over which f rises and where it falls;

(ii) the local extrema.

(d) Use the sign pattern for f

00 (x) to determine.

(i) where the graph of f is concave up and where it is concave down;

(ii) the inflection point(s) (if any). [20]

4. A rectangular sheet of metal of perimeter 36cm and dimensions x and y is to be rolled into a

cylinder. (Volume of cylinder is V = πr²h.) What values of x and y give the largest volume?

(The thickness of the metal can be neglected.)

Answer 1

$\displaystyle (1)\\ f(x) = \frac{e^{\frac{x}{2}}}{x}\\ (a)\\ f(0) = \infty, \\ \textsf{No}\,\, y-\textsf{intercept}\\ (b)\\ \textsf{vertical assumptote does not exist since}\\ e^{x/2} = 0, x \to -\infty\\ f(\infty) = 0\,\,\textsf{Horizontal assumptote is}\,\, y = 0.\\ (c)\\ f'(x) = \frac{1}{2}\frac{e^{\frac{x}{2}}}{x} - \frac{e^{\frac{x}{2}}}{x^2}\\ f(x)\,\,\textsf{rises if}\,\, f'(x) > 0\\ \implies \frac{xe^{\frac{x}{2}} - 2e^{\frac{x}{2}}}{2x^2} > 0\\ \implies e^{\frac{x}{2}}(x - 2) > 0 \\ \implies x > 2\\ f(x)\,\,\textsf{falls}\,\, f'(x) < 0\\ \implies \frac{xe^{\frac{x}{2}} - 2e^{\frac{x}{2}}}{2x^2} < 0\\ \implies e^{\frac{x}{2}}(x - 2) < 0 \\ \implies x < 2\\ \textsf{Local extrema}\,\, f(2) = \frac{e}{2}\\ (d)\\ f(x)\,\,\, \textsf{is concave up, iff}\,\,\,f''(x) > 0\\ \implies f(x) = \frac{2x^2e^{\frac{x}{2}} + x^3e^{\frac{x}{2}} - 2x^2e^{\frac{x}{2}} - 4x(x - 2)e^{\frac{x}{2}}}{4x^4} > 0\\ \implies x^2 - 4x + 8 > 0\\ \textsf{The roots of the quadratic}\\ \textsf{equation are complex}\\ \textsf{Hence, the function is not concave up}\\ f(x)\,\,\, \textsf{is concave down, iff}\,\,\,f''(x) < 0\\ \implies f(x) = \frac{2x^2e^{\frac{x}{2}} + x^3e^{\frac{x}{2}} - 2x^2e^{\frac{x}{2}} - 4x(x - 2)e^{\frac{x}{2}}}{4x^4} < 0\\ \implies x^2 - 4x + 8 < 0\\ \textsf{The roots of the quadratic}\\ \textsf{equation are complex}\\ \textsf{Hence, the function is not concave down}\\ \textsf{There is inflexion point iff}\,\,\,f''(x) = 0\\ \textsf{But, since the roots of the}\\ \textsf{quadratic equation are complex,}\\ \textsf{There are no inflexion points}\\ (2)\\\textsf{Area of the rectangle} = xy\\ \textsf{Volume of cylinder} = \pi r^2 h\\ \textsf{A larger surface area gives a}\\ \textsf{larger volume.}\\ 2(x + y) = 36, \, x + y = 18\\ A = x(18 - x)\\ A = x(18 - x)\\ A'(x)= 18 - 2x\\ A \,\, \textsf{is maximum or minimum at}\,\, A'(x) = 0\\ \implies 18 - 2x = 0, x = 9\\ y = 18 - x = 18 - 9 = 9\\ x = 9cm \,\, \textsf{and}\,\, y = 9cm\,\, \textsf{gives the largest volume.}$