Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 11

f(x) =x^4 - 2x^2 + 3
(a) To solve, take the derivative of the given function.
f'(x) =4x^3 - 4x
Then, set the derivative equal to zero.
0=4x^3-4x
Factor the right side of the equation.
0=4x(x^2-1)
0=4x(x-1)(x+1)
Set each factor equal to zero. And isolate the x in each equation.
Factor 1:
4x=0
x=0
Factor 2:
x-1=0
x=1
Factor 3:
x + 1 =0
x=-1
So, f(x) has three critical numbers. These are x=-1, x=0 and x=1. The intervals formed by these three critical numbers are:
(-oo, -1) (-1,0) (0,1) and (1,oo) .
To determine which among the intervals is the function increasing or decreasing, assign a test value for each. Plug-in the test value to the derivative
f'(x) = 4x^3-4x.
If the result of f'(x) is negative, the function is decreasing in that interval. And if the result is positive, the function is increasing in that interval.
For the first interval (-inf,-1), let the test value be x=-2.f'(-2) = 4(-2)^3 - 4(-2)=-24 (Decreasing)
For the second interval (-1,0), let the test value be x=-0.5 .f'(-0.5) = 4(-0.5)^3-4(-0.5)=1.5 (Increasing)
For the third interval (0,1), let the test value be x=0.5 .f'(0.5)=4(0.5)^3-4(0.5)=-1.5 (Decreasing)
And for the fourth interval (1, inf), let the test value be x=2.f'(2)=4(2)^3-4(2)=24 (Increasing)
Therefore, the function is increasing at (-1,0) uu (1, oo) and it is decreasing at (-oo,-1) uu (0,1) .
(b) For this part, let's refer to the change of signs of f'(x) before and after the critical number.
If f'(x) changes from positive to negative, then f(x) has a local maximum at x=c. If f'(x) changes from negative to positive, then f(x) has a local minimum at x=c.
For our critical numbers x=-1 and x=1, the change of signs of f'(x) before and after each critical number is from negative to positive. So, the local minimum of the function occurs at x=-1 and x=1. And the value of the local minimum is:
f(x)=x^4 -2x^2+3
f(-1)=(-1)^4-2(-1)^2+3=2
f(1)=(1)^4-2(1)^2+3=2
For our critical number x=0, the change of signs of f'(x) is from positive to negative. So, the local maximum of the function occurs at x=0. And the value of the local maximum is:
f(0) = 0^4-2(0)+3=3
Thus, f(x) has a local minimum of 2, which occurs at x=-1 and x=1. And, it has a local maximum of 3, which occurs at x=0.
(c) To solve, use the second derivative of the function.
f'(x)=4x^3-4x
f''(x)=12x^2-4
Set the second derivative equal to zero. And, solve for x.
0=12x^2-4
0=3x^2-1
1=3x^2
1/3=x^2
+-sqrt3/3=x
So, change of concavity occurs at x=-sqrt3/3 and x=sqrt3/3 . The intervals formed by these two inflections are:
(-oo, -sqrt3/3) (-sqrt3/3, sqrt3/3) and (sqrt3/3,oo) .
To determine at what intervals is the function concave upward or downward, assign a test value for each. And plug-in them to the second derivative
f''(x) = 12x^2-4
If the result of f"(x) is positive, the function is concave upward in that interval. If the result is negative, the function is concave downward in that interval.
For the first interval (-oo, -sqrt3/3) , let the test value be x=-1.
f''(-1)=12(-1)^2-4=8 (concave upward)
For the second interval (-sqrt3/3,sqrt3/3) , let the test value be x=0.
f''(0)=12(0)^2-4=-4 (concave downward)
And for the third interval (sqrt3/3,oo) , let the test value be x=1.
f''(1)=12(1)^2-4=8 (concave upward).
Hence, the function is concave upward at intervals (-oo,-sqrt3/3) uu (sqrt3/3,oo) . And it is concave downward at interval (-sqrt3/3,sqrt3/3) .
To solve for the inflection point, plug-in x=-sqrt3/3 and x=sqrt3/3 to the original function.
f(x)=x^4-2x^2+3
f(-sqrt3/3)=(-sqrt3/3)^4-2(-sqrt3/3)^2+3 =22/9
f(sqrt3/3)=(sqrt3/3)^4-2(sqrt3/3)^2+3=22/9
Therefore, the inflection points are (-sqrt3/3,22/9) and (sqrt3/3,22/9) .