Geometry and Calculus | Questions-Answers

Running Head: GEOMETRY
0
Geometry and Calculus
(Student Details: )
3/28/2020

GEOMETRY
1
Answer 1
Given function: f(x)=2x3+3x2−12x
(a) critical numbers are (−1,20) and (1,−7) out of which, (−1,20) is the maximum and
(1,−7) is a minimum.
It can be explained as below:
Now, while differentiating with respect to x' we have found:
f'(x)=6x2+6x−12 [1]
if we find value of above function at a critical point, f'(x)=0
f'(x)=0⇒6x2+6x−12=0
x2+x−2=0
(x+2)(x−1)=0
x=−2,1
Now, finding y-coordinate will require substituting these values into f(x)
x=−2⇒f(−2)=2(−8)+3(4)−12(−2)=−16+12+24=20
x=1⇒f(1)=2+3−12=−7
So the critical points are (−1,20) and (1,−7)
(b) As per the given function f(x)= 2x3+3x2−12x
In order to find the critical numbers, first we found values of x st f'(x)=0
Thus, we have determined the intervals of increasing and decreasing of the given function.
(c ) To find intervals of concavity, we have found where the plotted curve of the given
function is concave up or concave down. Thus, intervals of concavity are as follows:

GEOMETRY
2
When we differentiate equation [1] with respect to x ;
f''(x)=12x+6
x=−2⇒f''(−2)=−14+6<0, hence it is a maximum
x=1⇒f''(1)=12+6>0, hence it is a minimum
Concave down on (-∞, -1/2) since f”(x) is negative
Concave up on (-1/2, ∞) since f”(x) is positive
(d) To determine local maximum or minimum values, we have found where first derivative
became 0, thus:
(-2, 20) is a local maxima
(1,-7) is a local minima
(e ) given function is 2x3+3x2−12x
Thus, asymptotes are x (2x2+3x-12)
(f) Graph