Question

Let \(f\left( x \right) = {x^2}\).

a) Estimate the values \(f'\left( 0 \right),{\rm{ }}f'\left( {\frac{1}{2}} \right),{\rm{ }}f'\left( 1 \right),\) and \(f'\left( 2 \right)\) by zooming in on the graph of \(f\).

b) Use symmetry to deduce the values of \(f'\left( { - \frac{1}{2}} \right),{\rm{ }}f'\left( { - 1} \right),\) and \(f'\left( { - 2} \right).\)

c) Use the results from parts (a) and (b) to guess a formula for \(f'\left( x \right)\).

d) Use the definition of derivative to prove that your guess in part (c) is correct.

Short answer

a) The obtained values are \(f'\left( 0 \right) = 0,{\rm{ }}f'\left( {\frac{1}{2}} \right) = 1,{\rm{ }}f'\left( 1 \right) = 2,{\rm{ }}f'\left( 2 \right) = 4\).

b) The obtained values are \(f'\left( { - \frac{1}{2}} \right) = - 1\), \(f'\left( { - 1} \right) = - 2\), \(f'\left( { - 2} \right) = - 4\).

c) The formula for the \(f'\left( x \right)\) is \(2x\).

d) It is proved that the guess in part (c) is correct.

Step-by-step solution

Estimate the values of \(f'\) by zooming in on the graph of \(f\)

a) The procedure to draw the graph of the equation by using the graphing calculator is as follows:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\({X^2}\)in the\({Y_1}\)tab.
2. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the function \(f\left( x \right) = {x^2}\) by zooming in as shown below:

Use the graph of \(f\) to estimate the values of \(f'\left( 0 \right),{\rm{ }}f'\left( {\frac{1}{2}} \right),{\rm{ }}f'\left( 1 \right),{\rm{ }}f'\left( 2 \right)\) as shown below:

\(\begin{aligned}{l}f'\left( 0 \right) = 0\\f'\left( {\frac{1}{2}} \right) = 1\\f'\left( 1 \right) = 2\\f'\left( 2 \right) = 4\end{aligned}\)

Thus, the values are \(f'\left( 0 \right) = 0,{\rm{ }}f'\left( {\frac{1}{2}} \right) = 1,{\rm{ }}f'\left( 1 \right) = 2,{\rm{ }}f'\left( 2 \right) = 4\).

Use symmetry to deduce the value of \(f'\)

b) According to the symmetry \(f'\left( { - x} \right) = - f'\left( x \right)\) deduce the value of \(f'\) as shown below:

\(f'\left( { - \frac{1}{2}} \right) = - 1\)

\(f'\left( { - 1} \right) = - 2\)

\(f'\left( { - 2} \right) = - 4\)

Thus, the values are \(f'\left( { - \frac{1}{2}} \right) = - 1\), \(f'\left( { - 1} \right) = - 2\), \(f'\left( { - 2} \right) = - 4\).

Use the results from parts (a) and (b) to guess a formula for \(f'\left( x \right)\)

c) The explanation for the results of \(f'\left( x \right)\) from parts (a) and (b) is shown below:

\(f'\left( 0 \right) = 0,f'\left( {\frac{1}{2}} \right) = 1,f'\left( 1 \right) = 2,f'\left( 2 \right) = 4\)\( \Rightarrow \)The values of \(f'\left( x \right)\) is obtained by twice the value of \(x\).

\(f'\left( { - \frac{1}{2}} \right) = - 1\), \(f'\left( { - 1} \right) = - 2\), \(f'\left( { - 2} \right) = - 4\)\( \Rightarrow \) The values of \(f'\left( x \right)\) is obtained by twice the value of \(x\).

It is observed that \(f'\left( x \right)\) is twice the value of \(x\). Therefore, we might assume that \(f'\left( x \right) = 2x\).

Thus, the formula for the \(f'\left( x \right)\) is \(2x\).

Prove the guess in part (c) is correct

d) The derivative of the function \(f\) at any number \(a\), represented by \(f'\left( a \right)\),

\(f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\)

When this limit exists.

Evaluate the derivative of the function as shown below:

\(\begin{aligned}f'\left( x \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^2} - {x^2}}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{x^2} + 2hx + {h^2}} \right) - {x^2}}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{2hx + {h^2}}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {2x + h} \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \left( {2x + h} \right)\\ & = 2x\end{aligned}\)

Thus, it is proved that the guess in part (c) is correct.