Questions: For f(x)=x^4-4x^3+5 find the following. (A) f'(x) (B) The slope of the graph of f at x=-4 (C) The equation of the tangent line at x=-4 (D) The value(s) of x where the tangent line is horizontal (A) f'(x)= (B) At x=-4, the slope of the graph of f is . (C) At x=-4, the equation of the tangent line is y= . (D) The tangent line is horizontal at x= . (Use a comma to separate answers as needed.)

Solution Steps

Given \( f(x) = x^4 - 4x^3 + 5 \), we need to find the derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(x^4 - 4x^3 + 5) = 4x^3 - 12x^2 \]

To find the slope of the graph of \( f \) at \( x = -4 \), we substitute \( x = -4 \) into \( f'(x) \).

\[ f'(-4) = 4(-4)^3 - 12(-4)^2 = 4(-64) - 12(16) = -256 - 192 = -448 \]

First, find the point on the function at \( x = -4 \):

\[ f(-4) = (-4)^4 - 4(-4)^3 + 5 = 256 + 256 + 5 = 517 \]

Using the point-slope form of the equation of a line \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point:

\[ y - 517 = -448(x + 4) \]

Simplifying:

\[ y = -448x - 1792 + 517 = -448x - 1275 \]

The tangent line is horizontal where the derivative \( f'(x) \) is zero:

\[ 4x^3 - 12x^2 = 0 \]

Factoring out \( 4x^2 \):

\[ 4x^2(x - 3) = 0 \]

Setting each factor to zero gives:

\[ 4x^2 = 0 \quad \text{or} \quad x - 3 = 0 \]

\[ x = 0 \quad \text{or} \quad x = 3 \]

Final Answer