Answer to Question #143735 in Calculus for Kate

Question

Let E(t) be the number of errors made by a resident from the start of a shift until t hours into the shift. The instantaneous rate of change of errors made is E′(t) = t3 − 3 t2 + 2. (i) Sketch E(t). Label any minima, maxima and/or inflection points. On the same axes, draw a line tangent to E(t) at t∗. Label the coordinates of the intersection (be careful that your scales for the two functions match).

Answer 1

$$Solution$$

With the help of the graph of the function, we can find the extreme points and the inflection points of the function.

Now, we will find the critical point by equating E'(t) to zero, as follows:

$E'(t)=0\\ t^3-3t^2+2=0\\ \implies (t-1)(t^2-2t-2)=0\\$

For $(t-1)=0;$

$$t-1=0;\ t=1\\$$

For $(t^2-2t-2)=0$

$$t^2-2t-2=0\ ;\ t=1+\sqrt 3 ,\ t=1-\sqrt 3$$

Hence,

$$t=1, \\ t=1+\sqrt 3 , or\ 2.73\\ t=1-\sqrt 3,\ or\ -0.73$$

Now, we will find the second derivative.

$E''(t)=\frac{\delta }{\delta t}(t^3-3t^2+2)\\ \implies 3t^2-6t$

Now, the value of the second derivative at the critical point will be:

$E''(1)=3(1)^2-6(1)=-3 <0, \implies maximum\ point\\ E''(2.73)=3(2.73)^2-6(2.73)=6 <0, \implies minimum\ point\\ E''(-0.73)=3(-0.73)^2-6(-0.73)=6 <0, \implies minimum\ point\\$

Now, we have to find the inflection point, so for that, we have to put

$E''(t)=0\\ \implies 3t^2-6t=0\\ \implies t=0, t=2$

So, we have two inflection points.

So finally, the curve is: