Find all the critical numbers for the function f(x) = cube root of 9-x^2.
help please?!
y = (9-x^2)^(1/3)
there are two x=intercepts, (let y=0)
(3,0) and (-3,0)
there is a y-intercept at (0,9^(1/3)) , (let x=0)
dy/dx = (1/3)(9-x^2)^(-2/3)(-2x)
= (-2/3)(x)(9-x^2)^(-2/3)
= 0 for a max/min
x = 0
so (0,9^(1/3)) is a turning point, and has to be a maximum, (see below)
the graph is above the x-axis between -3 and 3
and below the x-axis for all other values of x
It is symmetric about the y-axis
The definition of the "critical points" of a function at this website might be helpful:
http://www.analyzemath.com/calculus/applications/critical_numbers.html
To find the critical numbers of the function f(x) = cube root of (9 - x^2), we need to find the values of x for which the derivative of the function is equal to zero or undefined.
First, let's find the derivative of the function f(x).
f(x) = (9 - x^2)^(1/3)
Using the chain rule, we can find the derivative:
f'(x) = (1/3)(9 - x^2)^(-2/3) * (-2x)
Simplifying the derivative:
f'(x) = (-2x) / (3 * (9 - x^2)^(2/3))
To find the critical numbers, we need to set the derivative equal to zero:
(-2x) / (3 * (9 - x^2)^(2/3)) = 0
Setting the numerator equal to zero, we have -2x = 0, which gives x = 0.
Now, we need to examine where the function is undefined. Since f(x) involves the cube root, the expression inside the cube root cannot be negative.
9 - x^2 >= 0
This simplifies to x^2 <= 9. Taking the square root, we have x <= 3 and x >= -3.
Combining both conditions, the critical numbers for the function f(x) = cube root of (9 - x^2) are x = -3, 0, and 3.
Therefore, the critical numbers are -3, 0, and 3.
To find the critical numbers of the function f(x) = ∛(9 - x^2), we need to determine where the derivative of the function is equal to zero or undefined.
First, let's find the derivative of f(x) using the chain rule.
f(x) = ∛(9 - x^2)
Let u = (9 - x^2)
f(x) = u^⅓
Applying the chain rule, the derivative is:
f'(x) = (u^⅓)' * (du/dx)
= (1/3) * u^(-2/3) * (-2x)
= (-2x) / (3∛(9 - x^2))^2/3
= -2x / (3∛(9 - x^2))^2/3
Now, we can set f'(x) equal to zero and solve for x:
-2x / (3∛(9 - x^2))^2/3 = 0
Since the numerator is zero, we have:
-2x = 0
Solving for x, we find:
x = 0
Next, we need to determine if there are any points where the derivative is undefined. The only way the derivative can be undefined is if the denominator equals zero:
3∛(9 - x^2) = 0
Cubing both sides, we have:
27(9 - x^2) = 0
9 - x^2 = 0
x^2 = 9
x = ±3
Therefore, the critical numbers of the function f(x) = ∛(9 - x^2) are x = 0, x = 3, and x = -3.