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

Solution Steps

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

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

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

(C) To find the equation of the tangent line at $x = -3$, 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 $(-3, f(-3))$.

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

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

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

Final Answer