Question

given (f(x)=3 - 5x^{3}), use a table to estimate the slope of the tangent line to (f) at the point (p(0,3)). 1. find the slope of the secant line (pq) for each point (q(x,f(x))) with the (x) - values given in the table. (round each answer to 6 decimal places if necessary.) 2. use the answers from the table to estimate the value of the slope of the tangent line at the point (p). (round your answer to the nearest integer.) provide your answer below:

(x)	(-0.01)	(-0.001)	(0.001)	(0.01)
(m_{pq})

Explanation:

Step1: Recall slope - formula for secant line

The slope of the secant line passing through $P(0,3)$ and $Q(x,f(x))$ is given by $m_{PQ}=\frac{f(x)-f(0)}{x - 0}=\frac{f(x)-3}{x}$. Given $f(x)=3 - 5x^{3}$, then $f(0)=3-5\times0^{3}=3$.

Step2: Calculate slope for $x = 0.01$

Substitute $x = 0.01$ into the slope formula:
\[

$$\begin{align*} m_{PQ}&=\frac{(3 - 5\times(0.01)^{3})-3}{0.01}\\ &=\frac{3-5\times10^{-6}-3}{0.01}\\ &=\frac{- 5\times10^{-6}}{0.01}\\ &=-0.0005 \end{align*}$$

\]

Step3: Calculate slope for $x = 0.001$

Substitute $x = 0.001$ into the slope formula:
\[

$$\begin{align*} m_{PQ}&=\frac{(3 - 5\times(0.001)^{3})-3}{0.001}\\ &=\frac{3-5\times10^{-9}-3}{0.001}\\ &=\frac{-5\times10^{-9}}{0.001}\\ &=-0.000005 \end{align*}$$

\]

Step4: Calculate slope for $x=-0.001$

Substitute $x=-0.001$ into the slope formula:
\[

$$\begin{align*} m_{PQ}&=\frac{(3 - 5\times(-0.001)^{3})-3}{-0.001}\\ &=\frac{3 + 5\times10^{-9}-3}{-0.001}\\ &=\frac{5\times10^{-9}}{-0.001}\\ &=0.000005 \end{align*}$$

\]

Step5: Calculate slope for $x=-0.01$

Substitute $x=-0.01$ into the slope formula:
\[

$$\begin{align*} m_{PQ}&=\frac{(3 - 5\times(-0.01)^{3})-3}{-0.01}\\ &=\frac{3+5\times10^{-6}-3}{-0.01}\\ &=\frac{5\times10^{-6}}{-0.01}\\ &=0.0005 \end{align*}$$

\]

Step6: Estimate slope of tangent line

As $x$ approaches $0$, the slopes of the secant lines approach the slope of the tangent line at $x = 0$. The average of the values as $x$ approaches $0$ from both sides suggests that the slope of the tangent line at $x = 0$ is $0$.

Answer:

For $x = 0.01$, $m_{PQ}=-0.0005$; for $x = 0.001$, $m_{PQ}=-0.000005$; for $x=-0.001$, $m_{PQ}=0.000005$; for $x=-0.01$, $m_{PQ}=0.0005$. The slope of the tangent line at $P(0,3)$ is $0$.