Calculus Mathematics Solution 2022

Running head: CALCULUS 1
Calculus
Student Name
Institution

CALCULUS 2
Question 1
Step 1: Identify the critical points by equating the derivative to zero
F(x) = x3+x2-5x+2
dy
dx = 3 x2 +2x-5
3 x2 +2x-5 = 0
Use the equation:
X=−b ± √b2−4 ac
2a = −2± √ 22−4∗3∗−5
2∗3 =−5
3 , 1
Step 2: find the domain
Since this is a quadratic equation, the domain is: −∞ <¿x¿ ∞
Step 3: check the behavior of the monotone intervals
Since the critical points fall within the domain, we form the monotone intervals as:
−∞<¿X← 5
3 , −5
3 <x<1, 1<x<∞
For −∞<¿x← 5
3 ;
We pick a value within the range (x=-1) and check the sign value of dy
dx
F(x) = 3(−1)2 +2(-1)-5=-4
For −5
3 <x<1;
We pick a value within the range (x=0) and check the sign value of dy
dx
F(x) = 3(0)2 +2(0)-5=-5
We pick a value within the range (x=2) and check the sign value of dy
dx

CALCULUS 3
F(x) = 3(2)2 +2(2)-5=11
−∞<¿x← 5
3
−5
3 <x<1 1<x<∞
Sign - - +
Behavior decreasing Decreasing Increasing
Question 2
Step 1: Find the first derivative
f(x) = x−2
x+ 1
Using quotient rule of differentiation;
d
dx ( x −2 ) ( x +1 ) − d
dx ( x +1 ) ( x −2 )
( x +1)2
= 1 ( x +1 ) −1( x−2)
(x +1)2 = 3
( x+1 )2
f ‘(x) = 3
( x+1 )2
Step 1: equate first derivative to zero and find where the first derivative is undefined
3
( x+1 ) 2 =0
The function 3
( x+1 )2 is undefined at x=-1 since we cannot divide by zero
Step 2: finding the domain
Domain: x<-1 or x>-1
Step 3: finding singularity points
x¿-1
Since f(x) is not defined at x¿-1, then no inflection points. Inflection points exists where a
function, f(x) is defined at all points within the domain (Hayter, 2013; Stephens, 2011).

CALCULUS 4
Question 3
Question 4
Step 1: Perimeter of the fence:
L+W = 80
Step 2: Making L the subject of the formula
L = 80-w
Step 3: Calculate the area
Area = L*W
= (80-W)W