Asked by Kate
Find all the critical numbers for the function f(x) = cube root of 9-x^2.
help please?!
help please?!
Answers
Answered by
Reiny
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
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
Answered by
drwls
The definition of the "critical points" of a function at this website might be helpful:
http://www.analyzemath.com/calculus/applications/critical_numbers.html
http://www.analyzemath.com/calculus/applications/critical_numbers.html
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