Question

Find all the maxima and minima of the
function f given by
f(x)= integration of [3(t-2)(t-3)^3 + 2(t-2)^2 (t-3)^2] dt. In the limit 2 to x

Answer 1

$$f(x)=\displaystyle\int_{2}^x[3(t-2)(t-3)^3+2(t-2)^2(t-3)^2]dt$$

Domain: $\R$

$$f'(x)=[3(x-2)(x-3)^3+2(x-2)^2(x-3)^2]\cdot1-0=$$

$$=(x-2)(x-3)^2(3x-9+2x-4)=(x-2)(x-3)^2(5x-13)$$

Find the critical number(s):

$$f'(x)=0=>(x-2)(x-3)^2(5x-13)=0$$

$$x_1=2, x_2={13 \over 5}, x_3=3$$

First Derivative Test

$$If\ x<2, f'(x)>0, f(x)\ increases$$

$$If\ 2<x<2.6, f'(x)<0, f(x)\ decreases$$

$$If\ 2.6<x<3, f'(x)>0, f(x)\ increases$$

$$If\ x>3, f'(x)>0, f(x)\ increases$$

$$\int[3(t-2)(t-3)^3+2(t-2)^2(t-3)^2]dt=$$

$$=3\int(t-3)^4dt+3\int(t-3)^3dt+$$

$$+2\int(t-3)^4dt+4\int(t-3)^3dt+2\int(t-3)^2dt=$$

$$=(t-3)^5+{7 \over 4}(t-3)^4+{2 \over 3}(t-3)^3+C$$

$$f(x)=\displaystyle\int_{2}^x[3(t-2)(t-3)^3+2(t-2)^2(t-3)^2]dt=$$

$$=(x-3)^5+{7 \over 4}(x-3)^4+{2 \over 3}(x-3)^3-{1 \over 12}$$

$$f(2)=0$$

$$f(2.6)=-0.09144$$

The function f(x) has a local maximum with value of at $x=2.$

The function f(x) has a local minimum with value of $(-0.09144)$ at $x=2.6.$