what is the relative maximum and minimum of the function.
f(x)=x^3+6x^2-36
I'll do this one, and maybe you can show where you get stuck on the others, if you do. If you just want confirmation of your answers, just show what you get.
max and min occur where f'(x) = 0
f'(x) = 3x^2 + 12x
= 3x(x+4)
f'=0 when x=0 or x = -4.
f(-4) = -4
f(0) = -36
Now, the question is, which is max, which is min?
Knowing what you do about the shape of cubics, it should be clear which is which.
The definitive way is to check the second derivative.
f''(x) = 6x+12
If f'' < 0, it's a max
If f'' > 0, it's a min.
Check it out, and you're done.
I don't understand
ok. I see this is for an algebra class, not calculus. In that case, things are a bit less straightforward. What is the subject of your current chapter? Unless you have some numerical tools at your command, I don't know of a general way to find maxima and minima for cubics and higher-order.
dividing polynomials possible answers are
max(-6,216)minimum(2,-40)
(-6,40) (2,-216)
(6,216) (-2,-40)
(6,40) (-2,-216)
oh, well. If they give you choices, just plug in the x values and see what pops out. However, the choices given do not appear to me to be correct for that polynomial.
visit wolframalpha.com and type in your function. It will show a graph, and you can see that the relative max and min are as I calculated above.
To find the relative maximum and minimum of a function, we need to find the critical points of the function and check their nature using the first and second derivative tests. Here's how to do it step by step for the given function f(x) = x^3 + 6x^2 - 36:
Step 1: Find the first derivative of the function.
The first derivative of f(x) = x^3 + 6x^2 - 36 is found by taking the derivative of each term with respect to x.
f'(x) = 3x^2 + 12x
Step 2: Find the critical points.
Critical points occur where the first derivative is equal to zero or undefined. Setting f'(x) = 0 and solving for x gives us:
3x^2 + 12x = 0
Factor out 3x to get:
3x(x + 4) = 0
So, we have two critical points:
x = 0 and x = -4
Step 3: Find the second derivative of the function.
To determine the nature of the critical points, we need to find the second derivative of the function.
Taking the derivative of f'(x) = 3x^2 + 12x, we get:
f''(x) = 6x + 12
Step 4: Evaluate the second derivative at the critical points.
Plug in the values of the critical points we found earlier into the second derivative:
f''(0) = 6(0) + 12 = 12
f''(-4) = 6(-4) + 12 = -12
Step 5: Determine the nature of the critical points.
If the second derivative is positive at a critical point, it implies a relative minimum, and if it's negative, it implies a relative maximum.
For x = 0, f''(0) = 12, which is positive. So, there is a relative minimum at x = 0.
For x = -4, f''(-4) = -12, which is negative. So, there is a relative maximum at x = -4.
Therefore, the function f(x) = x^3 + 6x^2 - 36 has a relative minimum at x = 0 and a relative maximum at x = -4.