The equation of the tangent line to the curve $f ( x ) = \left( 2 x ^ { 5 } - 6 x ^ { 2 } + 6 \right) \left( x ^ { 3 } + x - 1 \right)$ at the point (0, -6) is y = 6x - 6.

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To find the equation of the tangent line to the curve at a given point, we first need to find the derivative of the function, which gives us the slope of the tangent line at any point on the curve. The function given is $f(x) = (2x^5 - 6x^2 + 6)(x^3 + x - 1)$. We need to apply the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.Let's denote $u(x) = 2x^5 - 6x^2 + 6$ and $v(x) = x^3 + x - 1$. Then, $u'(x) = 10x^4 - 12x$ and $v'(x) = 3x^2 + 1$. Applying the product rule:$$f'(x) = (10x^4 - 12x)(x^3 + x - 1) + (2x^5 - 6x^2 + 6)(3x^2 + 1)$$To find the slope of the tangent line at $x = 0$, we substitute $x = 0$ into $f'(x)$:$$f'(0) = (10(0)^4 - 12(0))(0^3 + 0 - 1) + (2(0)^5 - 6(0)^2 + 6)(3(0)^2 + 1) = 6$$So, the slope of the tangent line at the point (0, -6) is 6. The equation of a line is given by $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. Since the line passes through (0, -6), when $x = 0$, $y = -6$, which means $b = -6$. Therefore, the correct equation of the tangent line is $y = 6x - 6$.Upon reevaluation, the initial conclusion was incorrect; the statement is actually true. The correct equation of the tangent line, given the slope and the point it passes through, is indeed $y = 6x - 6$.

The Equation of the Tangent Line to the Curve f(x)=(2x5−6x2+6)(x3+x−1)f ( x ) = \left( 2 x ^ { 5 } - 6 x ^ { 2 } + 6 \right) \left( x ^ { 3 } + x - 1 \right)f(x)=(2x5−6x2+6)(x3+x−1)

The Equation of the Tangent Line to the Curve $f ( x ) = \left( 2 x ^ { 5 } - 6 x ^ { 2 } + 6 \right) \left( x ^ { 3 } + x - 1 \right)$