How do I answer this calculus question step by step?

Question

For the function 𝑔(𝑥) = 𝑥⁴ − 8𝑥2 find all critical points and say which are local maximums or minimums.

Answer 1

While it is tempting to just dive in taking first and second derivatives, if we factor g(x) first, we can get a lot of understanding before taking any derivatives.

x^4 - 8x^2 = x^2(x^2 - 8) = x^2(x - √8)(x+√8)

We thus see that this function is even and we have a double root at 0, a root at -√8, and another root at √8

As the leading exponent is even and positive, we know that the function will go to ∞ as x goes to ±∞

Thus, the function will be positive on (-∞, -√8 ) U (√8, ∞ ) and negative on (√8, 0 ) U (0, √8 )

We will have local minima somewhere on (√8, 0 ) and (0, √8 ) and a local maximum at 0.

Now, let's take the first derivative and find the 2 local minima.

f'(x) = 4x^3 - 16x = 4x (x^2 - 4) = 4x (x-2) (x+2)

As we have already shown, the local maximum is at 0, and we will have local minima at ±2

(y = 0 when x = 0; y = 2^4 - 8 * 2^2 = -16 when x = 2 and (-2)^4 - 8 * (-2)^2 = -16 when x = -2)

We can show that there is a local max. at x = 0 and local minima at ±2 also by taking the second derivative.

f''(x) = 12x^2 - 16

f''(0) = -16 < 0, so this is a maximum.

f''(2) = 12(2^2) - 16 = 32 and f''(-2) = 12((-2)^2) - 16 = 32, so these are both minima.

Not asked, but there are 2 inflection points: f''(x) = 0 when 12x^2 - 16 = 0 or 12x^2 = 16 or x = 2√3/3.

Loooking at the roots of the graph, these are just where we would expect them to be, in (-2, 0) and (0, 2), symmetric about the y-axis.