To solve the given problems, we will follow these steps:
(A) To find f′(x) f'(x) f′(x), we need to differentiate the function f(x)=x4−12x3+3 f(x) = x^4 - 12x^3 + 3 f(x)=x4−12x3+3 with respect to x x x.
(B) To find the slope of the graph of f f f at x=−3 x = -3 x=−3, we will evaluate the derivative f′(x) f'(x) f′(x) at x=−3 x = -3 x=−3.
(C) To find the equation of the tangent line at x=−3 x = -3 x=−3, we will use the point-slope form of a line. We need the slope from part (B) and the point on the graph, which is (−3,f(−3)) (-3, f(-3)) (−3,f(−3)).
To find the derivative f′(x) f'(x) f′(x) of the function f(x)=x4−12x3+3 f(x) = x^4 - 12x^3 + 3 f(x)=x4−12x3+3, we differentiate each term with respect to x x x: f′(x)=4x3−36x2 f'(x) = 4x^3 - 36x^2 f′(x)=4x3−36x2
To find the slope of the graph of f f f at x=−3 x = -3 x=−3, we substitute x=−3 x = -3 x=−3 into the derivative: f′(−3)=4(−3)3−36(−3)2=−432 f'(-3) = 4(-3)^3 - 36(-3)^2 = -432 f′(−3)=4(−3)3−36(−3)2=−432
The equation of the tangent line can be found using the point-slope form of a line, y−y1=m(x−x1) y - y_1 = m(x - x_1) y−y1=m(x−x1), where m m m is the slope and (x1,y1)(x_1, y_1)(x1,y1) is the point on the graph. First, we find the y-coordinate of the point on the graph: f(−3)=(−3)4−12(−3)3+3=408 f(-3) = (-3)^4 - 12(-3)^3 + 3 = 408 f(−3)=(−3)4−12(−3)3+3=408 The slope m m m is −432-432−432, and the point is (−3,408)(-3, 408)(−3,408). Thus, the equation of the tangent line is: y−408=−432(x+3) y - 408 = -432(x + 3) y−408=−432(x+3) Simplifying, we get: y=−432x−888 y = -432x - 888 y=−432x−888
(A) The derivative is f′(x)=4x3−36x2 \boxed{f'(x) = 4x^3 - 36x^2} f′(x)=4x3−36x2.
(B) The slope of the graph at x=−3 x = -3 x=−3 is −432 \boxed{-432} −432.
(C) The equation of the tangent line at x=−3 x = -3 x=−3 is y=−432x−888 \boxed{y = -432x - 888} y=−432x−888.
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