
Gerome S.
asked 11/16/22Curve 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?
1 Expert Answer
A point of inflection occurs when the second derivative changes sign (from positive to negative, or vice-versa). Here, the second derivative is constantly 2, so it never changes sign, so there are no points of inflection.
Graphically, a point of inflection is when the curve switches between concave-up and concave-down. Your curve is a parabola, which is always concave-up, so there are no inflection points.
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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/dryl2slyfc11/16/22