What is the inverse of D/(Dx)8x^3-15x^2-2,x>= 1.5 ?

The inverse of D/(Dx)8x^3-15x^2-2,x>= 1.5 is sqrt(-(x+2)/7),-sqrt(-(x+2)/7)

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inverse of D/(Dx)8x^3-15x^2-2,x>= 1.5

inverse

\frac{D}{D x} 8 x^{3} - 15 x^{2} - 2, x \geq 1.5

- \frac{x + 2}{7}, - - \frac{x + 2}{7}

Solution steps

y = \frac{D}{D x} \cdot 8 x^{3} - 15 x^{2} - 2

Replace

x

with

y

x = \frac{D}{Dy} \cdot 8 y^{3} - 15 y^{2} - 2

Solve for

y, x = \frac{D}{Dy} 8 y^{3} - 15 y^{2} - 2

- \frac{x + 2}{7}, - - \frac{x + 2}{7}

35 x + 68.784

f (x) = (e^{x} - e)^{\frac{1}{3}}

\frac{6 x + 5}{5 x - 8}

f (x) = 10 x^{3} + 6

f (x) = ln (- \frac{x - 5}{2}) + 1

What is the inverse of D/(Dx)8x^3-15x^2-2,x>= 1.5 ?
The inverse of D/(Dx)8x^3-15x^2-2,x>= 1.5 is sqrt(-(x+2)/7),-sqrt(-(x+2)/7)