The function ${f(x) = {4\over {x^2}}+ x}$ is defined on the domain ${x > 0, x \in \mathbb{R}}$, the set of real numbers.
Find the maximum and minimum values of ${f(x)}$ on the closed interval ${1 \le x \le 4}$.
So my understanding is that stationary points occur when ${f'(x) = 0}$
In a closed interval, the maximum and minimum values of a function are either a stationary point or at an end of an interval.
${f'(x) = {-8x^{-3}} + 1}$ for stationary points
If I was drawing a sketch of this function, I would factorise to find values for x that I would then plug the values into the original function to find y.
I would then draw a nature table to get the feel of the curve.
My questions are:
- How do I Factorise ${-8x^{-3} + 1}$ to find values for x
- How do I find the maximum and minimum values that incorporates the closed interval?
Best Answer
So for first question we have that
$$ -\frac{8}{x^3}+1 = 0 $$ $$ -\frac{8}{x^3} = -1 $$ $$ x^3 = 8 $$ $$ x = 2 $$
now plug this into you function and then check the endpoints of the interval, that is $x = 1$ and $x=4$.