1. compute the derivative \( f'(x) \) of \( f(x) = x - x^3 \) and then \( f'(1) \) is the slope
2. use the point slope-point formula for a line
3. see 'Solution Path'

1. \( f'(1) = -2 \)
2. \( y = -2 (x-1) \)
3. see 'Solution Path'

1. The slope at the point \( (1,0) \) is the value \( f'(1) \) of the derivative \( f' \) of \( y = f(x) \) at \( x = 1 \), i.e. \[ \begin{eqnarray*} f'(x) & = & \lim_{h \to 0} \frac{(x+h) - (x^3 + 3x^2 h + 3 xh^2 + h^3) -x + x^3}{h}\\ & = & \lim_{h \to 0} \left( 1 - 3x^2 - 3xh - h^2 \right) \, \, = \, \, 1 - 3 x^2 \quad \Longrightarrow \quad f'(1) = -2 \, . \end{eqnarray*} \]
2. The tangent at \( P(p_x, p_y) = P(1,0) \) is a line with slope \( m=-2 \) through \( P \), i.e. by virtue of the point slope formula, we have \[ y - p_y \, \, = \, \, m \cdot (x - p_x) \quad \Longrightarrow \quad y \, \, = \, \, -2 \cdot (x-1) \, . \]
3. The closer we zoom to the point of tangency the less is the observable difference between the graph of the function and the tangent line. This circumstance will later enable us to use the tangent at a point as a 'good' approximation of the function (i.e. replace the discussion of the behavior of the function close to the point of tangency by a, often more easy, discussion of the behavior of the tangent line).
    
   
   

1. On an interval \([x_1, x_2]\), the average rate of change is \( \Delta y = f(x_2) = f(x_1) \) over \( \Delta x = x_2 - x_1 \).
2. Inspect the avearge rate for \([10,50]\) and generalize your findings.
3. Compute and then compare the two rates.
4. For which interval can this be interpreted as the arverage rate of change?

1. \(10\)
2. The average rate is \( 0 \) for all intervals \([x_1, x_2]\) with \(f(x_1) = f(x_2)\).
3. The interval \( [40,60] \) gives the larger average rate of change of the two.
4. This is the arverage rate of change on the interval \( [10, 40] \)

1. With \( f(20) = 300 \) and \( f(60) = 700 \) we have that the average rate of change is \[ \frac{\Delta y }{\Delta x} \, \, = \, \, \frac{f(60) - f(20)}{60-20} \, \, = \, \, \frac{400}{40} \, \, = \, \, 10 \,. \]
2. The average rate is \( 0 \) for all intervals \([t_1, t_2]\) with \(f(t_1) = f(t_2)\). For instance, \([10,50]\) as \(f(10) = f(50) = 400\).
3. The interval \( [40,60] \) gives the larger average rate of change of the two, as \[ \begin{eqnarray*} \frac{f(60) - f(40)}{60-40} & = & \frac{700 - 200}{20} \, \, = \, \, 25 \\ \frac{f(70) - f(40)}{70-40} & = & \frac{900 - 200}{30} \, \, \approx \, \, 23.3333 \end{eqnarray*} \]
4. The quantity \[ \frac{f(40) - f(10)}{40 - 10} \, \, = \, \, \frac{200 - 400}{30} \, \, \approx \, \, -6.6667 \] gives the arverage rate of change on the interval \( [10, 40] \).

Recall, a function is not differentiable at those points, where it is not continuous and/ or where the limit of the difference quotient does not exist. Where do such cases arise for the given function?

• A function is not differentiable at those points, where it is not continuous. Here, this is the case at \( x = -1 \).
• A function is not differentiable at those points, where the limit of the difference quotient does not exist, i.e. \[ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \, \, \, \text{does not exist} \, . \] This can happen at 'kinks' at which the one-sided limits are different. Here, at \(x = 2\) there is such a kink.

Successively make use of the given information to get a rough sketch of the behavior of the graph and then 'smoothen' your result. I.e.
 

...

We see

• horizontal tangents at \( (-3,2) \), \( (0,-2) \), and \( (3,2) \), therefore \( f' \) vansihes for \( x \in \{-3, 0, 3\} \).
• \( f \) is strictly decreasing on \( (-3,0) \) and strictly increasing in \( (0, 3) \), therefore \( f' \) is negative for \( x \in (-3,0) \) and positive for \( x \in (0,3) \).
• by means of slope triangles we estimate the value of the slope at different points of the graph.

Apply the differentiation rules such that you assume that \( x \) is the variable for differentiation and \( t \) is a constant. Then, apply the differentiation rules such that you assume that \( t \) is the variable for differentiation and \( x \) is a constant.

The derivative w.r.t. \( x \) is \[ \frac{\textrm{d} \, f(x,t)}{\textrm{d} \, x} \, \, = \, \, f_x(x,t) \, \, = \, \, \frac{\textrm{d}}{\textrm{d} \, x} \left( \frac{t}{x^2} \, + \frac{x}{t} \right) \, \, = \, \, -\frac{2 t}{x^3} \, + \, \frac{1}{t} \, \, = \, \, \frac{x^3 - 2t^2}{x^3 t} \, , \] and the derivative w.r.t. \( t \) is \[ \frac{\textrm{d} \, f(x,t)}{\textrm{d} \, t} \, \, = \, \, f_t(x,t) \, \, = \, \, \frac{\textrm{d}}{\textrm{d} \, t} \left( \frac{t}{x^2} \, + \frac{x}{t} \right) \, \, = \, \, \frac{1}{x^2} \, - \, \frac{x}{t} \, \, = \, \, \frac{t - x^3}{x^2 t} \, . \]