Final answer:
The equation of the normal line to the curve y=3x³-4x at the point (1,-1) is found by calculating the negative reciprocal of the derivative at that point. The derivative gives us a slope of 5, making the normal line's slope -1/5. Thus, the equation of the normal line is y = -1/5x - 4/5.
Step-by-step explanation:
The question is asking for the equation of the normal line to the curve y=3x³-4x at the point (1,-1). To find this equation, we first need to determine the slope of the tangent line to the curve at (1, -1) by finding the derivative of the given function.
The derivative of y=3x³-4x is y'=9x²-4. Plugging in the x-value of the point (1, -1), we find that the slope of the tangent line at that point is y'(1)=9(1)²-4=5. The slope of the normal line is the negative reciprocal of the tangent line's slope, which is -1/5.
To find the equation of the normal line, we use the point-slope form of a line, which is y-y1=m(x-x1), where (x1, y1) is the point on the line and m is the slope. Substituting the point (1, -1) and slope -1/5, the equation of the normal line is y - (-1) = -1/5(x - 1), which simplifies to y = -1/5x + 1/5 - 1, or y = -1/5x - 4/5.