Question

How do you graph `f(x)=1/x-3x^3` using the information given by the first derivative?

Answer 1

Look for zeros on the first and second derivatives.

Explanation:

Start by finding the first and second derivative.

`f(x)=1/x-3x^3=x^(-1)-3x^3`

rewrite `f(x)` such that it fits the power rule.

`f'(x)=-x^(-2)-9x^(2)=-1/(x^(2))-9x^2`

Now look at the first derivative: both `-1/x^2` and `- 9x^2` are negative for real values of x. Additionally, the first derivative is not defined for `x=0` and demonstrates asymptotic behavior for that value. Thus `f(x)` shall consist of two branches and given that `f'(x)<0`, `f(x)` is decreasing over both branches.

Zeros and signs on the second derivative give additional information about the concavity of and inflection points (if any) on `f(x)`.

`f''(x)=2x^(-3)-18x=2/(x^3)-18x`

Solving `f''(x)=0` gives

`x^4=1/9`,
`x=pm 1/sqrt(3)`

`f''(x)` shows asymptotic behavior at `x=0` so checking the value of `f''(x)` on the four open intervals

• `(-infty, -1/sqrt(3))`
• `(-1/sqrt(3), 0)`
• `(0, 1/sqrt(3))`, and
• `(1/sqrt(3), infty)`

shall give a comprehensive conclusion on the sign of `f''(x)` over the whole real domain.

• `f''(x)>0` for `x in (-infty, -1/sqrt(3))` and `x in (0, 1/sqrt(3))`;
• `f''(x)<0` for `x in (-1/sqrt(3),0)` and `x in (1/sqrt(3),0)`.

Conclusion:

• `f(x)` demonstrates asymptotic behavior on `x=0`
• `f(x)` is decreasing on both of its branches (from 1st derivative).

taking information from the second derivative into account:

• `f(x)` is concave upward on `x in (-infty, -1/sqrt(3))` and `x in (0, 1/sqrt(3))`

• `f(x)` is concave downward on `x in (-1/sqrt(3),0)` and `x in (1/sqrt(3),0)`.

• `f(x)` has inflection points at `x=pm 1/sqrt(3)`

FYI Here's a plot of `f(x)` graph{1/x-3x^3 [-1.5, 1.5, -5, 5]}