Question

(a) Sketch the graph of \(f\left( x \right) = {\bf{1}} + \sqrt {x + {\bf{3}}} \) by starting with the graph of \(y = \sqrt x \) and using the transformations of Section 1.3.

(b) Use the graph from part (a) to sketch the graph of \(f'\).

(c) Use the definition of derivative to find \(f'\left( x \right)\). What are the domain of \(f\) and \(f'\)?

(d) Graph \(f'\) and compare with your sketch in part (b).

Short answer

(a) The graph of the function is shown below:

(b) The graph of the derivative is shown below:

(c) The derivative is \(f'\left( x \right) = \frac{1}{{2\sqrt {x + 3} }}\). The domain of \(f\left( x \right)\) is \(\left( { - 3,\infty } \right)\) and domain of \(f'\left( x \right)\) is \(\left( { - 3,\infty } \right)\).

(d) The graph is shown below:

Step-by-step solution

Find the graph of the function \(f\left( x \right) = {\bf{1}} + \sqrt {x + {\bf{3}}} \)

a) Following transformations should be used to obtain the graph of \(y = 1 + \sqrt {x + 3} \) from the graph of \(y = \sqrt x \).

1. Shift the graph of \(y = \sqrt x \), 3 units towards the left.
2. Then shift the graph of \(y = \sqrt {x + 3} \), 1 unit upwards.

The figure below represents the curve of \(y = \sqrt x \).

The figure below represents the curve of \(y = \sqrt {x + 3} \).

The figure below represents the curve of \(y = 1 + \sqrt {x + 3} \).

Sketch the graph of \(f'\)

b) It can be observed from the graph of \(f\left( x \right)\) that slope is positive at all the points, but as \(x \to - {3^ + }\), then \(f'\left( x \right) \to \infty \).

Initially, the slope is decreasing at a faster rate, whereas as \(x\) is approaching to \(\infty \), the slope tends to zero.

The figure below represents the graph of \(f'\left( x \right)\).

Find the value of \(f'\left( x \right)\)

The derivative \(f'\left( x \right)\) can be calculated using the formula \(f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\).

\(\begin{aligned}f'\left( x \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {1 + \sqrt {\left( {x + h} \right) + 3} } \right) - \left( {1 + \sqrt {x + 3} } \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\sqrt {\left( {x + h} \right) + 3} } \right) - \left( {\sqrt {x + 3} } \right)}}{h} \times \frac{{\left( {\sqrt {\left( {x + h} \right) + 3} + \sqrt {x + 3} } \right)}}{{\left( {\sqrt {\left( {x + h} \right) + 3} + \sqrt {x + 3} } \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\left( {x + h} \right) + 3} \right) - \left( {x + 3} \right)}}{{h\left( {\sqrt {\left( {x + h} \right) + 3} + \sqrt {x + 3} } \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( {\sqrt {\left( {x + h} \right) + 3} + \sqrt {x + 3} } \right)}}\\ & = \frac{1}{{2\sqrt {x + 3} }}\end{aligned}\)

So, the derivative of the function is \(\frac{1}{{2\sqrt {x + 3} }}\).

Find the domain of \(f\left( x \right)\) and \(f'\left( x \right)\)

The function \(f\left( x \right) = 1 + \sqrt {x + 3} \) is defined, when;

\(\begin{aligned}{c}x + 3 \ge 0\\x \ge - 3\end{aligned}\)

The domain of the function is \(\left( { - 3,\infty } \right)\).

Similarly, the domain of the derivative \(f'\left( x \right) = \frac{1}{{2\sqrt {x + 3} }}\) is;

\(\begin{aligned}{c}x + 3 > 0\\x > - 3\end{aligned}\)

Thus, the domain of the derivative of \(f'\left( x \right)\) is \(\left( { - 3,\infty } \right)\).

Sketch the graph of \(f'\)

Use the following commands to obtain the graph of \(f'\left( x \right)\) by using the graphing calculator:

1. In the graphing calculator, select “STAT PLOT” and enter the equation \(1/\left( {2{{\left( {X + 3} \right)}^{1/2}}} \right)\) in the \({Y_1}\) tab.

Enter the graph button in the graphing calculator.

Thus, the sketch is obtained, and the sketch in part (b) is similar.