(1 point)
For what values of $x$ is the tangent line of the graph of
\[
f(x)=10 x^{3}+15 x^{2}-362 x+120
\]
parallel to the line $y=1.9-2 x$ ?
(If there are multiple values then separate them with commas.)
\[
x=
\]
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Final Answer: The values of \(x\) for which the tangent line of the graph of \(f(x)=10 x^{3}+15 x^{2}-362 x+120\) is parallel to the line \(y=1.9-2 x\) are \(\boxed{-4, 3}\).
Step 1 :The slope of the line \(y=1.9-2x\) is -2. The tangent line to the function \(f(x)\) will be parallel to this line when the derivative of \(f(x)\) is equal to -2.
Step 2 :We need to find the derivative of \(f(x)\), set it equal to -2, and solve for \(x\).
Step 3 :The derivative of \(f(x) = 10x^{3} + 15x^{2} - 362x + 120\) is \(f'(x) = 30x^{2} + 30x - 362\).
Step 4 :Setting \(f'(x)\) equal to -2 gives us the equation \(30x^{2} + 30x - 362 = -2\).
Step 5 :Solving this equation gives us the solutions \(x = -4, 3\).
Step 6 :Final Answer: The values of \(x\) for which the tangent line of the graph of \(f(x)=10 x^{3}+15 x^{2}-362 x+120\) is parallel to the line \(y=1.9-2 x\) are \(\boxed{-4, 3}\).