What are the points of inflection, if any, of #f(x)=-7x^3+27x^2-30x+1

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Question

What are the points of inflection, if any, of #f(x)=-7x^3+27x^2-30x+1

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Elijah Abram · Accepted Answer

second derivative vanishes gives
#-42x + 54 = 0#
or #x = 9/7# The third order derivative is an inflection point and is constant.

Charles Arocha · Answer

undefined To find the points of inflection, we need to find the second derivative of the function and then solve for where the second derivative equals zero. After finding those points, we can determine if they are points of inflection by checking the concavity of the function at those points.

First, let's find the second derivative of $ f(x) = -7x^3 + 27x^2 - 30x + 1 $:

$$ f''(x) = -42x + 54 $$

Next, let's set the second derivative equal to zero and solve for $ x $:

$$ -42x + 54 = 0 $$
$$ -42x = -54 $$
$$ x = \frac{54}{42} $$
$$ x = \frac{9}{7} $$

Now, we need to check the concavity of the function at $ x = \frac{9}{7} $.

$$ f''\left(\frac{9}{7}\right) = -42\left(\frac{9}{7}\right) + 54 = -54 + 54 = 0 $$

Since the second derivative changes sign at $ x = \frac{9}{7} $, this is a point of inflection.

Therefore, the point of inflection for the function $ f(x) = -7x^3 + 27x^2 - 30x + 1 $ is $ \left(\frac{9}{7}, f\left(\frac{9}{7}\right)\right) $.