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

Solution

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)) $.

Step 1: Differentiate the Function

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

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

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

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

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: \[ y = -432x - 888 \]