Question

ma 241 si: sections 3.1 and 3.2 - tangent lines and the derivative at a point & the derivative as a function
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1. the formula for the general derivative of a function ( f(x) ) with respect to ( x ) is ____________________. the derivative of a function ( f(x) ) evaluated at a specific point ( x = x_0 ) gives the ________________ rate of change of ( f(x) ) at ( x = x_0 ). the derivative of ( f ) at ( x = x_0 ) also gives the slope of the ________________ line to the curve ( f(x) ) at ( x = x_0 ).

follow these steps to find the equation of a tangent line to the curve ( f(x) ) at ( x = x_0 ):

1. find the slope of the graph of ( f(x) ) at ( x = x_0 ) by evaluating ________________.
2. if not already given, find the ( y )-coordinate that corresponds to ( x_0 ) by evaluating ________________.
3. plug the information found in the previous steps into the equation of a line. the point-slope form of the equation of a line is ______________ and the slope-intercept form is ______________.

1. given ( f(x) = x^2 - 4x - 4 ), answer the following questions.

a) find the general derivative ( f(x) ) using the steps below:
step 1: find ( f(x + h) ).
step 2: find ( f(x + h) - f(x) ).
step 3: find ( \frac{f(x+h)-f(x)}{h} ) and simplify your answer.
step 4: find ( lim_{h \to 0} \frac{f(x+h)-f(x)}{h} ).
b) now that ( f(x) ) has been found, find ( f(1) ). what does this value represent?

Explanation:

Part 1: Conceptual Fill - ins

The general formula for the derivative of a function \(y = f(x)\) with respect to \(x\) is given by the limit definition: \(f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}\). The derivative of a function at a point \(x = x_0\) gives the instantaneous rate of change of the function at that point. Also, the derivative at a point gives the slope of the tangent line to the curve at that point.

For finding the equation of the tangent line:

1. The slope of the graph of \(f(x)\) at \(x=x_0\) is found by evaluating \(f^{\prime}(x_0)\) (or using the limit definition \(\lim_{h

ightarrow0}\frac{f(x_0 + h)-f(x_0)}{h}\)).

1. The \(y\) - coordinate corresponding to \(x_0\) is found by evaluating \(f(x_0)\).
2. The point - slope form of a line is \(y - y_0=m(x - x_0)\) where \((x_0,y_0)\) is a point on the line and \(m\) is the slope. The slope - intercept form is \(y=mx + b\) where \(m\) is the slope and \(b\) is the \(y\) - intercept.

Part 2a: Finding the derivative of \(f(x)=x^{2}-4x - 4\) using the limit definition

Step 1: Find \(f(x + h)\)

We substitute \(x+h\) into the function \(f(x)\).
If \(f(x)=x^{2}-4x - 4\), then \(f(x + h)=(x + h)^{2}-4(x + h)-4\).
Expanding \((x + h)^{2}\) using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), we get \(f(x + h)=x^{2}+2xh+h^{2}-4x-4h - 4\).

Step 2: Find \(f(x + h)-f(x)\)

Substitute \(f(x + h)=x^{2}+2xh+h^{2}-4x-4h - 4\) and \(f(x)=x^{2}-4x - 4\) into \(f(x + h)-f(x)\):
\[

$$\begin{align*} f(x + h)-f(x)&=(x^{2}+2xh+h^{2}-4x-4h - 4)-(x^{2}-4x - 4)\\ &=x^{2}+2xh+h^{2}-4x-4h - 4 - x^{2}+4x + 4\\ &=2xh+h^{2}-4h \end{align*}$$

\]

Step 3: Find \(\frac{f(x + h)-f(x)}{h}\) and simplify

Divide \(f(x + h)-f(x)=2xh+h^{2}-4h\) by \(h\) (assuming \(h
eq0\)):
\[
\frac{f(x + h)-f(x)}{h}=\frac{2xh+h^{2}-4h}{h}=\frac{h(2x + h-4)}{h}=2x+h - 4
\]

Step 4: Find \(\lim_{h

ightarrow0}\frac{f(x + h)-f(x)}{h}\)
Take the limit as \(h
ightarrow0\) of \(\frac{f(x + h)-f(x)}{h}=2x+h - 4\):
\[
\lim_{h
ightarrow0}(2x+h - 4)=2x+0 - 4=2x - 4
\]
So, \(f^{\prime}(x)=2x - 4\)

Part 2b: Finding \(f^{\prime}(1)\) and its interpretation

Step 1: Evaluate \(f^{\prime}(x)\) at \(x = 1\)

Substitute \(x = 1\) into \(f^{\prime}(x)=2x - 4\):
\(f^{\prime}(1)=2(1)-4=2 - 4=-2\)

Interpretation:

The value \(f^{\prime}(1)=-2\) represents the slope of the tangent line to the curve \(y = f(x)=x^{2}-4x - 4\) at the point where \(x = 1\). It also represents the instantaneous rate of change of the function \(f(x)\) at \(x = 1\).

Final Answers

1. The formula for the general derivative of a function \(f(x)\) with respect to \(x\) is \(\boldsymbol{\lim_{h

ightarrow0}\frac{f(x + h)-f(x)}{h}}\). The derivative of a function \(f(x)\) evaluated at a specific point \(x = x_0\) gives the \(\boldsymbol{\text{instantaneous}}\) rate of change of \(f(x)\) at \(x = x_0\). The derivative of \(f\) at \(x = x_0\) also gives the slope of the \(\boldsymbol{\text{tangent}}\) line to the curve \(f(x)\) at \(x = x_0\).

• 1. Find the slope of the graph of \(f(x)\) at \(x = x_0\) by evaluating \(\boldsymbol{f^{\prime}(x_0)}\) (or \(\boldsymbol{\lim_{h

ightarrow0}\frac{f(x_0 + h)-f(x_0)}{h}}\)).

• 2. If not already given, find the \(y\) - coordinate that corresponds to \(x_0\) by evaluating \(\boldsymbol{f(x_0)}\).
• 3. The point - slope form of the equation of a line is \(\boldsymbol{y - y_0=m(x - x_0)}\) and the slope - intercept form is \(\boldsymbol{y=mx + b}\).

1. a)

• Step 1: \(f(x + h)=\boldsymbol{x^{2}+2xh+h^{2}-4x-4h - 4}\)
• Step 2: \(f(x + h)-f(x)=\boldsymbol{2xh+h^{2}-4h}\)
• Step 3: \(\frac{f(x + h)-f(…

Answer:

The general formula for the derivative of a function \(y = f(x)\) with respect to \(x\) is given by the limit definition: \(f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}\). The derivative of a function at a point \(x = x_0\) gives the instantaneous rate of change of the function at that point. Also, the derivative at a point gives the slope of the tangent line to the curve at that point.

For finding the equation of the tangent line:

1. The slope of the graph of \(f(x)\) at \(x=x_0\) is found by evaluating \(f^{\prime}(x_0)\) (or using the limit definition \(\lim_{h

ightarrow0}\frac{f(x_0 + h)-f(x_0)}{h}\)).

1. The \(y\) - coordinate corresponding to \(x_0\) is found by evaluating \(f(x_0)\).
2. The point - slope form of a line is \(y - y_0=m(x - x_0)\) where \((x_0,y_0)\) is a point on the line and \(m\) is the slope. The slope - intercept form is \(y=mx + b\) where \(m\) is the slope and \(b\) is the \(y\) - intercept.

Part 2a: Finding the derivative of \(f(x)=x^{2}-4x - 4\) using the limit definition

Step 1: Find \(f(x + h)\)

We substitute \(x+h\) into the function \(f(x)\).
If \(f(x)=x^{2}-4x - 4\), then \(f(x + h)=(x + h)^{2}-4(x + h)-4\).
Expanding \((x + h)^{2}\) using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), we get \(f(x + h)=x^{2}+2xh+h^{2}-4x-4h - 4\).

Step 2: Find \(f(x + h)-f(x)\)

Substitute \(f(x + h)=x^{2}+2xh+h^{2}-4x-4h - 4\) and \(f(x)=x^{2}-4x - 4\) into \(f(x + h)-f(x)\):
\[

$$\begin{align*} f(x + h)-f(x)&=(x^{2}+2xh+h^{2}-4x-4h - 4)-(x^{2}-4x - 4)\\ &=x^{2}+2xh+h^{2}-4x-4h - 4 - x^{2}+4x + 4\\ &=2xh+h^{2}-4h \end{align*}$$

\]

Step 3: Find \(\frac{f(x + h)-f(x)}{h}\) and simplify

Divide \(f(x + h)-f(x)=2xh+h^{2}-4h\) by \(h\) (assuming \(h
eq0\)):
\[
\frac{f(x + h)-f(x)}{h}=\frac{2xh+h^{2}-4h}{h}=\frac{h(2x + h-4)}{h}=2x+h - 4
\]

Step 4: Find \(\lim_{h

ightarrow0}\frac{f(x + h)-f(x)}{h}\)
Take the limit as \(h
ightarrow0\) of \(\frac{f(x + h)-f(x)}{h}=2x+h - 4\):
\[
\lim_{h
ightarrow0}(2x+h - 4)=2x+0 - 4=2x - 4
\]
So, \(f^{\prime}(x)=2x - 4\)

Part 2b: Finding \(f^{\prime}(1)\) and its interpretation

Step 1: Evaluate \(f^{\prime}(x)\) at \(x = 1\)

Substitute \(x = 1\) into \(f^{\prime}(x)=2x - 4\):
\(f^{\prime}(1)=2(1)-4=2 - 4=-2\)

Interpretation:

The value \(f^{\prime}(1)=-2\) represents the slope of the tangent line to the curve \(y = f(x)=x^{2}-4x - 4\) at the point where \(x = 1\). It also represents the instantaneous rate of change of the function \(f(x)\) at \(x = 1\).

Final Answers

1. The formula for the general derivative of a function \(f(x)\) with respect to \(x\) is \(\boldsymbol{\lim_{h

ightarrow0}\frac{f(x + h)-f(x)}{h}}\). The derivative of a function \(f(x)\) evaluated at a specific point \(x = x_0\) gives the \(\boldsymbol{\text{instantaneous}}\) rate of change of \(f(x)\) at \(x = x_0\). The derivative of \(f\) at \(x = x_0\) also gives the slope of the \(\boldsymbol{\text{tangent}}\) line to the curve \(f(x)\) at \(x = x_0\).

• 1. Find the slope of the graph of \(f(x)\) at \(x = x_0\) by evaluating \(\boldsymbol{f^{\prime}(x_0)}\) (or \(\boldsymbol{\lim_{h

ightarrow0}\frac{f(x_0 + h)-f(x_0)}{h}}\)).

• 2. If not already given, find the \(y\) - coordinate that corresponds to \(x_0\) by evaluating \(\boldsymbol{f(x_0)}\).
• 3. The point - slope form of the equation of a line is \(\boldsymbol{y - y_0=m(x - x_0)}\) and the slope - intercept form is \(\boldsymbol{y=mx + b}\).

1. a)

• Step 1: \(f(x + h)=\boldsymbol{x^{2}+2xh+h^{2}-4x-4h - 4}\)
• Step 2: \(f(x + h)-f(x)=\boldsymbol{2xh+h^{2}-4h}\)
• Step 3: \(\frac{f(x + h)-f(x)}{h}=\boldsymbol{2x+h - 4}\)
• Step 4: \(\lim_{h

ightarrow0}\frac{f(x + h)-f(x)}{h}=\boldsymbol{2x - 4}\) (so \(f^{\prime}(x)=2x - 4\))
b) \(f^{\prime}(1)=\boldsymbol{-2}\). This value represents the slope of the tangent line to the curve \(y = f(x)\) at \(x = 1\) (or the instantaneous rate of change of \(f(x)\) at \(x = 1\)).