Questions: For f(x)=x^4-12x^3+3 find the following. (A) f'(x) (B) The slope of the graph of f at x=-3 (C) The equation of the tangent line at x=-3 (D) The value(s) of x where the tangent line is horizontal

For f(x)=x^4-12x^3+3 find the following.
(A) f'(x)
(B) The slope of the graph of f at x=-3
(C) The equation of the tangent line at x=-3
(D) The value(s) of x where the tangent line is horizontal
Transcript text: For $f(x)=x^{4}-12 x^{3}+3$ find the following. (A) $f^{\prime}(x)$ (B) The slope of the graph of $f$ at $x=-3$ (C) The equation of the tangent line at $x=-3$ (D) The value(s) of $x$ where the tangent line is horizontal
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Solution

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Solution Steps

To solve the given problems, we will follow these steps:

(A) To find f(x) f'(x) , we need to differentiate the function f(x)=x412x3+3 f(x) = x^4 - 12x^3 + 3 with respect to x x .

(B) To find the slope of the graph of f f at x=3 x = -3 , we will evaluate the derivative f(x) f'(x) at x=3 x = -3 .

(C) To find the equation of the tangent line at 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)) .

Step 1: Differentiate the Function

To find the derivative f(x) f'(x) of the function f(x)=x412x3+3 f(x) = x^4 - 12x^3 + 3 , we differentiate each term with respect to x x : f(x)=4x336x2 f'(x) = 4x^3 - 36x^2

Step 2: Evaluate the Derivative at x=3 x = -3

To find the slope of the graph of f f at x=3 x = -3 , we substitute x=3 x = -3 into the derivative: f(3)=4(3)336(3)2=432 f'(-3) = 4(-3)^3 - 36(-3)^2 = -432

Step 3: Find the Equation of the Tangent Line at x=3 x = -3

The equation of the tangent line can be found using the point-slope form of a line, yy1=m(xx1) y - y_1 = m(x - x_1) , where m m is the slope and (x1,y1)(x_1, y_1) is the point on the graph. First, we find the y-coordinate of the point on the graph: f(3)=(3)412(3)3+3=408 f(-3) = (-3)^4 - 12(-3)^3 + 3 = 408 The slope m m is 432-432, and the point is (3,408)(-3, 408). Thus, the equation of the tangent line is: y408=432(x+3) y - 408 = -432(x + 3) Simplifying, we get: y=432x888 y = -432x - 888

Final Answer

(A) The derivative is f(x)=4x336x2 \boxed{f'(x) = 4x^3 - 36x^2} .

(B) The slope of the graph at x=3 x = -3 is 432 \boxed{-432} .

(C) The equation of the tangent line at x=3 x = -3 is y=432x888 \boxed{y = -432x - 888} .

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