Answer:

a) [-1, ∞-) and [1, ∞+)

b)  (√2/2,∞)  and (0,-√2/2)

c) y=0, y=2 and y=4

Step-by-step explanation:

a) On what intervals is f increasing?  

We need to proceed some steps, to answer it properly. What are the critical points? So  

1) Differentiate the original function

f(x)=3x^{5}-5x^3+2\ f'(x)=15x^4-15x^{2}

2) Turn this into an equation  

15x^4-15x^2=0\ u=x^{2} and \u^{2}=x^{4} \ Then\ 15u^{2} -15u=0\ \15u(u^2 -1)=0\Therefore\u=0,u=1

Substitute back to x

x^{2} =u\ x^{2} =u^4\ x^{2}=1\ x= 1,x=-1\ x^{2} =0\ x=0

Those are x-coordinates of the critical points of f(x)=3x^{5}-5x^3+2

Using the critical points x, to find y-coordinates of those critical points: namely, maximum point, saddle point and minimum point

x=-1 y=3x^{5}-5x^3+2

y=3(-1)^{5}-5(-1)^3+2

y=4

(-1,4) Maximum point

x=0 y=3x^{5}-5x^3+2

y=3(0)^{5}-5(0)^3+2 y=2

(0,2) Saddle Point

x=1 y=3x^{5}-5x^3+2

y=3(1)^{5}-5(1)^3+2  

(1,0) Minimum point

f(x) increases when any value of  x ∈ (-∞, -1] and x ∈ [1, ∞+)  is plugged in the function.  Or we can rewrite as  xxleq -1\ \xgeq 1

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b)On what intervals is the graph of f concave upward?

To find out which intervals are these, we need to calculate the 2nd derivative.

f(x)=3x^{5}-5x^3+2\ f(x)'=15x^4-15x^{2} \f(x)''=60x^3-30x\

Since we want the concave upward

f(x)”>0

Then:

60x^3-30x>0\ 30x(2x^2-1)>0\

Rewriting

30x(sqrt{2x}-1)

Then

x=0, x=frac{sqrt{2}}{2} \-frac{sqrt{2}}{2}

Studying the sign

The intervals >0, when the graph is concave upwards.

are (√2/2,∞)  and (0,-√2/2)

c) Write the equation of each horizontal tangent line to the graph of f.

Generally, every time somebody ask for a tangent line of any given function, they provide an information. A specific point.

But when they do not? Like in this case?

We have to look for where the 1st function’s derivative is zero.

As we ‘ve done previously in letter a. x=-1, x=0, x=1

Let’s plug them back to the original function

y=3x^{5}-5x^{3}+2\   y=3(-1)^{5}-5(-1)^{3}+2\ y=-3-5+2\ y=-8+2\ y=4\ \y=3x^{5}-5x^{3}+2\ y=3(0)^{5}-5(0)^{3}+2\ y=2\\ y=3x^{5}-5x^{3}+2\ y=3(1)^{5}-5(1)^{3}+2\ y=0

So each horizontal tangent line, tangents the maximum, saddle and the minimum point.