Gerome S.

asked • 11/16/22

Curve Tracing, Find critical Points and solve for points of inflection if any.

y = 4 − 6x + x2

so first is to find first derivative

y'=-6+2x

then y' = 0

0=-6+2x divide both side by 2 then

x=3

to find y substitute x to the equation

y= 4- 6(3) + (3)2

y= 4 - 18 + 9

y= -5

then test the critical points by getting 2nd derivative

f''(x) = 2

f''(3) = 2

theres no way to check since there's no x value to substitute for.

so how am i able to solve for Points of inflection, is there any solution or there is just no point of inflection?

Doug C.

The function is the equation of a parabola (upward opening since the coefficient of x^2 is positive). Since the 2nd derivative is always positive (a constant 2), the curve is concave upward everywhere. That means no points of inflection because the concavity never changes. Take a look here: desmos.com/calculator/dryl2slyfc
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11/16/22

Christina C.

tutor
Critical points come from the first derivative, Inflection points come from the second derivative. Since f’’(x) = # , that means the second derivative is constant (never changing) thus it is entirely concave up (since the # you solved for is positive) and it never changes, no inflection point. To check the critical points to be a max or min, you test out values before and after 3. (Say 2, f’(2) = -6 +(2)(2) = -6 + 4 = -2) thus (-infinity, 3) is decreasing since negative. Now test the other side, ( f’(4) = -6 + 2(4) = -6 + 8 = 2) thus (3, +infinity) is increasing since positive. So we go from decreasing on the left, to increasing on the right, thus we create x=3 a MINIMUM.
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11/16/22

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