Answer:
Step-by-step explanation:
a) you have that:
[tex]f'(x)=\frac{(x-2)(x+8)}{(x+1)(x-5)},\ \ x\neq -1,5[/tex]
The derivative of a function equals to zero allows you to find the critical points:
[tex]f'(x)=0\\\\(x-2)(x+8)=0\\\\x_1=2\\\\x_2=-8[/tex]
x=2,8 are the critical points
b) To know the behavior if f it is necessary to know where are f is indeterminated. The derivative give to you information about the slope of f. For x=-1,5 you have an infinite slope. Hence, for that values of x you have two indetermination s of f(x.)
However, if you see the atacched images you can obser that the original function (that is obtained with the intgegral) does not have available values for x<5 due to the logrithms. Hence, there are no critical points
The function only increases after x=5 from -infinity to +infinity
c) there are no local maximum neither local minimum
