Question

How do you sketch the graph `y=x^4+2x^2-3` using the first and second derivatives?

Answer 1

See the sketch of the graph below

Explanation:

We need

`(x^n)'=nx^(n-1)`

`a^2-b^2=(a-b)(a+b)`

Let `f(x)=x^4+2x^2-3`

We can factorise

`f(x)=(x^2+3)(x^2-1)=(x^2+3)(x+1)(x-1)`

`f'(x)=4x^3+4x=4x(x^2+1)`

We find critical points when `f'(x)=0`

`4x(x^2+1)=0` for `x=0`

We can calculate the second derivative

`f''(x)=12x^2+4`

`f''(0)=4 >0`, this is a minimum

We can do a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-1` `color(white)(aaaaaaaa)` `0` `color(white)(aaaaaaaa)` `1` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aaaa)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` ###color(white)(aaaa)#`↘` `color(white)(aaaaa)` `-3` `color(white)(aaaaaaaa)` `↗`

`lim_(x->+-oo)f(x)=+oo`

graph{(y-(x^4+2x^2-3))=0 [-7.9, 7.9, -3.95, 3.95]}