Asked by Reza
Find any critical numbers of the function. (Enter your answers as a comma-separated list.)
g(x) = x^7 − 7x^5
Honestly, I have no idea how to find the critical number!
g(x) = x^7 − 7x^5
Honestly, I have no idea how to find the critical number!
Answers
Answered by
Reiny
by critical numbers, they mean
the turning points, (maximim and minimum points)
any points of inflection
the x and the y -intercepts
so for y-intercept, let x = 0
so g(0) = 0, the origin (0,0) is both an x and a y intercept.
for more x-intercepts , let y = 0
x^7 - 7x^5 = 0
x^5(x^2 - 7) = 0
so x = 0 , x = ± √7
g'(x) = 7x^6 - 35x^4 = 0 for max/mins
7x^4(x^2 - 5) = 0
x = 0 or x = ± √5
g''(x) = 42x^5 - 140x^3
= 0 for any points of inflection
14x^3(3x^2 - 10) = 0
x = 0 or x = ± √(10/3)
since (0,0) is both a turning point and a point of inflection, It is neither a maximim nor a minimum
so you have 4 other different x values to sub back into the original equation to find the corresponding y value of the points,
I will leave that up to you
the turning points, (maximim and minimum points)
any points of inflection
the x and the y -intercepts
so for y-intercept, let x = 0
so g(0) = 0, the origin (0,0) is both an x and a y intercept.
for more x-intercepts , let y = 0
x^7 - 7x^5 = 0
x^5(x^2 - 7) = 0
so x = 0 , x = ± √7
g'(x) = 7x^6 - 35x^4 = 0 for max/mins
7x^4(x^2 - 5) = 0
x = 0 or x = ± √5
g''(x) = 42x^5 - 140x^3
= 0 for any points of inflection
14x^3(3x^2 - 10) = 0
x = 0 or x = ± √(10/3)
since (0,0) is both a turning point and a point of inflection, It is neither a maximim nor a minimum
so you have 4 other different x values to sub back into the original equation to find the corresponding y value of the points,
I will leave that up to you
Answered by
Reza
Thank you I really appreciate your help :)
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