Chapter 2, Problem 37RE

(a)

To determine

To find: The derivative of the function $f\left(x\right)=\sqrt{3-5x}$.

Answer to Problem 37RE

The derivative of $f\left(x\right)$ is $f^{\prime }\left(x\right)=-\frac{5}{2\sqrt{3-5x}}$.

Explanation of Solution

Formula used:

The derivative of a function f, denoted by $f^{\prime }\left(x\right)$, is

$f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ (1)

Calculation:

Obtain the derivative of the function $f\left(x\right)$.

Compute $f^{\prime }\left(x\right)$ by using the equation (1),

$\begin{array}{c}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\sqrt{3-5\left(x+h\right)}-\sqrt{3-5x}}{h}\end{array}$

Multiply and divide $\sqrt{3-5\left(x+h\right)}+\sqrt{3-5x}$ in the above expression

$\begin{array}{c}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{\sqrt{3-5\left(x+h\right)}-\sqrt{3-5x}}{h}\times \frac{\sqrt{3-5\left(x+h\right)}+\sqrt{3-5x}}{\sqrt{3-5\left(x+h\right)}+\sqrt{3-5x}}\\ =\underset{h\to 0}{\lim }\frac{{\left(\sqrt{3-5\left(x+h\right)}\right)}^2-{\left(\sqrt{3-5x}\right)}^2}{h\left\{\sqrt{3-5\left(x+h\right)}+\sqrt{3-5x}\right\}}\\ =\underset{h\to 0}{\lim }\frac{3-5\left(x+h\right)-\left(3-5x\right)}{h\left\{\sqrt{3-5\left(x+h\right)}+\sqrt{3-5x}\right\}}\\ =\underset{h\to 0}{\lim }\frac{-5h}{h\left\{\sqrt{3-5\left(x+h\right)}+\sqrt{3-5x}\right\}}\end{array}$

Since the limit h approaches zero but not equal to zero, cancel the common term h from both the numerator and the denominator,

$\begin{array}{c}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{-5}{\left\{\sqrt{3-5\left(x+h\right)}+\sqrt{3-5x}\right\}}\\ =\frac{-5}{\sqrt{3-5\left(x+0\right)}+\sqrt{3-5x}}\\ =-\frac{5}{2\sqrt{3-5x}}\end{array}$

Thus, the value of the derivative is $f^{\prime }\left(x\right)=-\frac{5}{2\sqrt{3-5x}}$.

(b)

To determine

To find: The domain of f and $f^{\prime }$

Answer to Problem 37RE

The domain of f is $\left(-\infty ,\frac{3}{5}\right]$ and the domain of $f^{\prime }$ is $\left(-\infty ,\frac{3}{5}\right)$.

Explanation of Solution

Calculation:

Obtain the domain of $f\left(x\right)=\sqrt{3-5x}$.

Since square root is only defined for non-negative integers,

So, take $3-5x\ge 0$ which implies $x\le \frac{3}{5}$.

Thus, the domain of f is $\left(-\infty ,\frac{3}{5}\right]$.

Obtain the domain of $f^{\prime }\left(x\right)=-\frac{5}{2\sqrt{3-5x}}$.

Since square root is only defined for non-negative integers and denominator is undefined at zero.

So, take $3-5x>0$ which implies $x<\frac{3}{5}$.

Thus, the domain of $f^{\prime }$ is $\left(-\infty ,\frac{3}{5}\right)$.

(c)

To determine

To check: The answer to part (a) is reasonable by comparing the graphs of $f$ and $f^{\prime }.$

Answer to Problem 37RE

The answer to part (a) is reasonable.

Explanation of Solution

Graph:

Use the online graphing calculator to draw the graph of $f\left(x\right)$ and $f^{\prime }\left(x\right)$ as shown below in Figure 1.

Observation:

From Figure 1, it is observed that the slope of the function $f\left(x\right)=\sqrt{3-5x}$

is always negative.

Also, the derivative graph $f^{\prime }\left(x\right)=-\frac{5}{2\sqrt{3-5x}}$ has negative functional value everywhere.

Thus, it can be concluded that the answer to part (a) is reasonable.