What is the inflection f(x)=2x^3-3x^2+7x-4 ?

The inflection f(x)=2x^3-3x^2+7x-4 is (1/2 ,-1)

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inflection f(x)=2x^3-3x^2+7x-4

inflection points

f (x) = 2 x^{3} - 3 x^{2} + 7 x - 4

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

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} + 7 x - 4 : - 1

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

4 x - 3 y = 9

y = 6 x - 2

5 x + 8

f (x) = - 4 - x^{2}

f (x) = \frac{2 x}{x + 4}

What is the inflection f(x)=2x^3-3x^2+7x-4 ?
The inflection f(x)=2x^3-3x^2+7x-4 is (1/2 ,-1)