Question

(2) For the function f(x) = x^2/3 - x^2 - 24x + 19 (i) Determine the turning points and the values of the function at each turning point. (ii) Determine the point of inflection, any exist. (iii) What is the range of value over which the function is strictly increasing (iv) What is the range of values over which the function is concave upwards. (3) A production manager determines that t months after production on a new product begins, the number of units produced will be N thousand, where N(t) = 17t^2 + 13t/(2r + 1)^2 . (i) What would be the level of production 5 months into operations. (ii) What would be the level of production be in the long run.

Answer 1

2 (i) Turning points

find where f'(x) is equal to zero or undefined

$f'(x)=x^2-2x-24$

Critical points x=-4 and x=6

$plug \ x=-4 \ into \ \frac{x^3}{3}-x^2-24x+19\\=\frac{233}{3}$

Maximum turning point $(-4, \frac{233}{3})$

$plug \ x=6 \ into \ \frac{x^3}{3}-x^2-24x+19\\=-89$

minimum turning point $(6,-89)$

Turning points $=(-4, \frac{233}{3})$, $(6,-89)$

(ii) Inflection point

find where f''(x) is equal to zero or undefined

$f''(x)=2x-2$

$x=1$

$plug \ x=1 \ into \ \frac{x^3}{3}-x^2-24x+19\\=\frac{-17}{3}$

inflection point $=(1, \frac{-17}{3})$

(iii) Increasing range on $(-\infin, -4), (6,\infin)$

(iv) Concave up range on $(1,\infin)$

3(i) level of production in 5 months

$\lim _{t\to 5}(17t^2+\frac{13t}{(2t+1)^2})$

$=(17\times5^2+\frac{13\times5}{(2\times5+1)^2})$

$N(t)=425.53719\\\\level \ of production \ will\ be =425537.19\ units$

(ii) find the limit

$\lim _{t\to \infin}(17t^2+\frac{13t}{(2t+1)^2})$

$=\lim _{t\to \infin}(17t^2)+\lim _{t\to \infin}(\frac{13t}{(2t+1)^2})$

$\lim _{t\to \infin}(17t^2)=\infin$

$\lim _{t\to \infin}(\frac{13t}{(2t+1)^2})=0$

$\infin +0=\infin$

In the long run the number of units produced approaches infinity.