THANK YOU TUTORS SO MUCH FOR YOUR HELP

Determine the intervals on which the function is concave up or concave down.

I finally understand that when f'' is greater than 0 it means concave up.

f(X) = 18x^2 + x^4

so I differentiate that to be
36x + 4x^3 that is f'
if I differentiate again I get:
f'' = 36+12x^2

it's from here that I get lost. At what intervals does it represent concave up or down? and what is my next step after getting f''?

look at f" - it is always positive

so, f is always concave up

just looking at it you can see that f(0) = 0, and always goes up on each side.

Right, but I am unsure what to put for intervals of concave up :/

how about (-oo,+oo) ?

To determine the intervals on which the function is concave up or concave down, you need to analyze the concavity of the function by examining the sign of the second derivative, f''.

In your case, you have found that f'' = 36 + 12x^2.

To identify the intervals where the function is concave up or concave down, follow these steps:

1. Start by finding the critical points of the function. Critical points occur where the second derivative is equal to zero or undefined. In this case, since f'' is a polynomial, it is defined for all values of x. Therefore, we do not have any critical points.

2. To determine the intervals of concavity, analyze the sign of the second derivative, f''. When f'' > 0, the function is concave up. When f'' < 0, the function is concave down.

To find the intervals where f'' > 0, solve the inequality f'' > 0:

36 + 12x^2 > 0

Simplifying the inequality, we get:

12x^2 > -36

Divide both sides by 12 to isolate x^2:

x^2 > -3

Since x^2 cannot be negative, the inequality holds true for all real values of x. Therefore, f'' > 0 for all x.

This means that the function f(x) = 18x^2 + x^4 is concave up on its entire domain.

So, in summary, the function f(x) = 18x^2 + x^4 is concave up for all values of x. There are no intervals of concave down for this particular function.