Question

Find all local extremums of the function, using both the First Derivative Test and Second Derivative Test

Answer 1

Answer: -4, -2, 4, and 16. Take my answer with a huge grain of salt. Im am prolly wrong

Step-by-step explanation:

You want to find x values where the derivative will be undefined or equal to 0. This will give us the critical points of our function. We're gonna have to do the quotient rule here. There might be another way, but im more confident using the quotient rule:

$\frac{f'(x)*g(x)-f(x)*g'(x)}{g(x)^2}$

So lets find f prime and g prime. f' just equals -1 and g' is solved using the power rule.

f'(x) = -1

g'(x) = 2x + 4

Your next step is plugging in everything to the quotient rule.

$\frac{(-1)(x^2+4x+5)-(7-x)(2x+4)}{(x^2+4x+5)^2}$

Unfortunately Im going to do this on paper because itll take too long for me to write it out.

$f'(x)=\frac{x^2-14x-33}{x^4+16x^2+25}$ is what I got. Notice how we can make both of these equal to 0 because the numerator will truly give us 0, and the denominator will give us undefined. Thats what we need to find and get those critical points

Im also going to do this on paper because it will take me too long to write it out. Essentially, make them both equal to zero. The first one is non factorable, which is weird imo. Im getting around -2 and +16 when using the quadratic formula. For the 2nd, I got -4 and +4