What is the inflection f(x)=2x^3-3x^2+6x-7 ?

The inflection f(x)=2x^3-3x^2+6x-7 is (1/2 ,-9/2)

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inflection f(x)=2x^3-3x^2+6x-7

inflection points

f (x) = 2 x^{3} - 3 x^{2} + 6 x - 7

(\frac{1}{2}, - \frac{9}{2})

Solution steps

f^{''} (x) = 12 x - 6

Find intervals:

Concave Downward

: - \infty < x < \frac{1}{2},

Concave Upward

: \frac{1}{2} < x < \infty

Plug

x = \frac{1}{2}

into

2 x^{3} - 3 x^{2} + 6 x - 7 : - \frac{9}{2}

(\frac{1}{2}, - \frac{9}{2})

1 + \frac{5}{x} \cdot \frac{5}{x}

f (x) = (x - 1)^{3}

1 - 2 x

f (x) = \frac{x ^{2}}{4 x - 3}

(- 1, 6), (0, 1)

What is the inflection f(x)=2x^3-3x^2+6x-7 ?
The inflection f(x)=2x^3-3x^2+6x-7 is (1/2 ,-9/2)