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Consider the following. f(x) = x^3 - 7x^2 Find f'(x). f'(x) = 3x^2 - 14x Solve f'(x) = 0 for x. (Enter your answers as a comma-separated list.) x = 0, 14/3 Find the points on the graph of f where the tangent line is horizontal. smaller x-value (x, y) = ( larger x-value (x, y) = ( 

          Consider the following.
f(x) = x^3 - 7x^2
Find f'(x).
f'(x) = 3x^2 - 14x
Solve f'(x) = 0 for x. (Enter your answers as a comma-separated list.)
x = 0, 14/3
Find the points on the graph of f where the tangent line is horizontal.
smaller x-value (x, y) = (
larger x-value (x, y) = (
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Consider the following. f(x) = x^3 - 7x^2 Find f'(x). f'(x) = 3x^2 - 14x Solve f'(x) = 0 for x. (Enter your answers as a comma-separated list.) x = 0, 14/3 Find the points on the graph of f where the tangent line is horizontal. smaller x-value (x, y) = ( larger x-value (x, y) = (

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Consider the following. f(x) = x^3 - 7x^2 Find f'(x). f'(x) = 3x^2 - 14x Solve f'(x) = 0 for x. (Enter your answers as a comma-separated list.) x = 0, 14/3 Find the points on the graph of f where the tangent line is horizontal. smaller x-value (x, y) = larger x-value (x, y) =
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Transcript

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00:01 Right, for this problem, we are giving the following function.
00:03 Here we are given the function f of x, which equals x to the power of 3 minus 7x to the power of 2.
00:10 Next, we are going to find the derivative.
00:13 So the derivative f prime of x is equal to 3x squared minus 14th x by using the power rule for derivatives.
00:24 Next, we want to solve for the equation setting the derivative.
00:30 Equal to 0 when setting the derivative equal to 0 we get 3x squared minus 14 x set this equal to 0 and we can do that by factoring out um x so x times 3x minus 14th equal 0 set each of the factors equal to 0 so we get that x equals 0 or 3x minus 14th equal 0 so in the second case we get that x equals 14th over 3 so the two values that we get is that x is either 0 or 14th over 3...
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