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What are the inflection points for #f(x) = -x^4-9x^3+2x+4#?

Answer 1

Charles Arocha

f'(x)= #-4x^3 -27x^2 +2#

f"(x)= #-12x^2-54x#

Inflection points would be given by f''(x)=0, that is #-12x^2-54x#=0,

=-6x( 2x+9)=0 x= 0, #-9/2# are the inflection points.

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Answer 2

Isabella Ackerman

To find the inflection points, we first need to find the second derivative of the function and then find where it equals zero or is undefined. The second derivative of (f(x) = -x^4 - 9x^3 + 2x + 4) is (f''(x) = -12x^2 - 54x + 2). Setting (f''(x) = 0) gives ( -12x^2 - 54x + 2 = 0), which doesn't factor nicely. Using the quadratic formula, we get (x = \frac{-(-54) \pm \sqrt{(-54)^2 - 4(-12)(2)}}{2(-12)}). Solving this gives (x \approx -4.26) and (x \approx 1.68). So, the inflection points are approximately ((-4.26, f(-4.26))) and ((1.68, f(1.68))).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.