Question

Use the given graph to estimate the value of each derivative. Then sketch the graph of \(f'\).

1. (a) \(f'\left( { - 3} \right)\) (b) \(f'\left( { - 2} \right)\) (c) \(f'\left( { - 1} \right)\) (d) \(f'\left( 0 \right)\)

(e) \(f'\left( 1 \right)\) (f) \(f'\left( 2 \right)\) (g) \(f'\left( 3 \right)\)

Short answer

The value of \(f'\left( { - 3} \right) \approx - 1\), \(f'\left( { - 2} \right) \approx 0\), \(f'\left( { - 1} \right) \approx \frac{1}{2}\), \(f'\left( 0 \right){\kern 1pt} \approx \frac{3}{2}\), \(f'\left( 1 \right) \approx 3\), \(f'\left( 2 \right) \approx 0\), \(f'\left( 3 \right) \approx - \frac{3}{2}\).

Step-by-step solution

The slope of tangent line 

We can approximate the value of the derivative at any value of \(x\) by forming tangent at the point \(\left( {x,f\left( x \right)} \right)\) and predicting its slope. The value of slope is the value of derivative of the function at that point.

1. From the given graph, it can be observed that the rise and run for \(x = - 3\) is approximately the same and at \(x = - 3\), \(f\left( x \right)\) is decreasing. So, the approximate value of the slope of the function at \(x = - 3\) is \( - 1\) means \(f'\left( { - 3} \right) = - 1\).

2. Similarly, the tangent at \(x = - 2\) is a horizontal line, therefore the slope is zero means \(f'\left( { - 2} \right) = 0\).

3.The rise is less than the run at \(x = - 1\) and at \(x = - 1\), \(f\left( x \right)\) is increasing. So, the approximate value of the slope of the function \(x = - 1\) is \(\frac{1}{2}\) means \(f'\left( { - 1} \right) = \frac{1}{2}\).

4. The rise is greater than run for \(x = 0\) and at \(x = 0\), \(f\left( x \right)\) is increasing. So, the approximate value of the slope of the function at \(x = 0\) is \(\frac{3}{2}\) means \(f'\left( 0 \right) = \frac{3}{2}\) .

5. The rise is three times the run for \(x = 1\) and at \(x = 1\), \(f\left( x \right)\) is increasing. So, the approximate value of the slope of the function at \(x = 1\) is \(3\) means \(f'\left( 1 \right) = 3\) .

6. The tangent at \(x = 2\) is a horizontal line, therefore the slope at \(x = 2\) is zero means \(f\left( 2 \right) = 0\).

7. The rise is greater than run for \(x = 3\) and at \(x = 3\), \(f\left( x \right)\) is decreasing. So, the approximate value of the slope of the function at \(x = 3\) is \( - \frac{3}{2}\) means \(f'\left( 3 \right) = - \frac{3}{2}\).

Draw the graph using the points

Plot the points \(\left( { - 3, - 1} \right),\left( { - 2,0} \right),\left( { - 1,\frac{1}{2}} \right),\left( {0,\frac{3}{2}} \right),\left( {1,3} \right),\left( {2,0} \right),\left( {3, - \frac{3}{2}} \right)\) on the graph and connect them to draw the curve of \(f'\left( x \right)\) as follows: