inflection point at (2,3).
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A critical point is a point where the derivative of the function equals 0. So, f'(1)=0.
Differentiating f(x) with respect to x yields 3x^2+2ax+b, so substituting 1 for x yields 3+2a+b=0.
An inflection point is where the second derivative of the function equals 0, so f''(2)=0.
Differentiating the derivative of f(x) yields 6x+2a, so substituting 2 produced 12+2a=0, or a=-6.
Plugging that into the first equation, 3+2a+b=0 makes 3+2(-6)+b=0, or b=9. But you still don't have y.
Here you can use the fact that the critical point or the inflection point have to be on the function, so f(1)=5 or f(2)=3. The first one's slightly easier to calculate, so we'll use that one. f(1)=1^3+(-6)(1)^2+9(1)+y=5, or 1-6+9+y=5, y=1. So you get a=-6, b=9, and y=1. Sorry if the way I explained it was overly complicated, feel free to contact me if you're still confused.
Differentiating f(x) with respect to x yields 3x^2+2ax+b, so substituting 1 for x yields 3+2a+b=0.
An inflection point is where the second derivative of the function equals 0, so f''(2)=0.
Differentiating the derivative of f(x) yields 6x+2a, so substituting 2 produced 12+2a=0, or a=-6.
Plugging that into the first equation, 3+2a+b=0 makes 3+2(-6)+b=0, or b=9. But you still don't have y.
Here you can use the fact that the critical point or the inflection point have to be on the function, so f(1)=5 or f(2)=3. The first one's slightly easier to calculate, so we'll use that one. f(1)=1^3+(-6)(1)^2+9(1)+y=5, or 1-6+9+y=5, y=1. So you get a=-6, b=9, and y=1. Sorry if the way I explained it was overly complicated, feel free to contact me if you're still confused.
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f(1) = 5
f'(1) = 0
f(2) = 3
f''(2) = 0
Four pieces of information, but only three parameters? Hmmm. Cross your fingers!
That f''(2) = 0 tells you that 6(2) + 2a = 0, so a = -6
That f'(1) = 0 and a=-6 tells you that 3(1)^2 + 2(-6)(1) + b = 0, so b=9.
That f(1) = 5, a=-6, b=9 tells you that 1^3 + (-6)(1)^2 + (9)(1)+y=5, so y=1
Finally, as a check compute f(2) if a=-6, b=9, y=1: 2^3 + (-6)(2)^2 + (9)(2)+1 = 3, which happily agrees with the claim that (2, 3) lies on the curve.
f'(1) = 0
f(2) = 3
f''(2) = 0
Four pieces of information, but only three parameters? Hmmm. Cross your fingers!
That f''(2) = 0 tells you that 6(2) + 2a = 0, so a = -6
That f'(1) = 0 and a=-6 tells you that 3(1)^2 + 2(-6)(1) + b = 0, so b=9.
That f(1) = 5, a=-6, b=9 tells you that 1^3 + (-6)(1)^2 + (9)(1)+y=5, so y=1
Finally, as a check compute f(2) if a=-6, b=9, y=1: 2^3 + (-6)(2)^2 + (9)(2)+1 = 3, which happily agrees with the claim that (2, 3) lies on the curve.