Question

instructions

1. work out this problem on one (1) sheet of plain paper. label is fe work out #1.
2. scan this sheet together with the other workout questions and submit in canvas, in the final exam workout questions portal.

work - out #1
(10 pts) use the definition of the derivative: $f(x)=limlimits_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$, to compute the derivative of $f(x)=4x - x^{2}$. no credit for correct answer without correct calculus (mathematical) work.
(5 pts) find the equation of the tangent line to the graph of the function at $x = 3$. express the equation in slope - intercept form: $y=mx + b$.
click to proceed when you have completed the work on your own paper.

Explanation:

Step1: Compute \( f(x + h) \)

Given \( f(x)=4x - x^{2} \), substitute \( x + h \) into \( f \):
\( f(x + h)=4(x + h)-(x + h)^{2}=4x + 4h-(x^{2}+2xh + h^{2})=4x + 4h - x^{2}-2xh - h^{2} \).

Step2: Compute \( f(x + h)-f(x) \)

Subtract \( f(x) \) from \( f(x + h) \):
\( f(x + h)-f(x)=(4x + 4h - x^{2}-2xh - h^{2})-(4x - x^{2})=4h - 2xh - h^{2} \).

Step3: Compute the difference quotient \( \frac{f(x + h)-f(x)}{h} \)

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

Step4: Take the limit as \( h\to0 \)

Find \( \lim_{h\to0}\frac{f(x + h)-f(x)}{h} \):
\( \lim_{h\to0}(4 - 2x - h)=4 - 2x \). So, \( f'(x)=4 - 2x \).

Step5: Find the slope at \( x = 3 \)

Substitute \( x = 3 \) into \( f'(x) \):
\( f'(3)=4 - 2(3)=4 - 6=-2 \). This is the slope \( m \) of the tangent line.

Step6: Find the point on the graph at \( x = 3 \)

Compute \( f(3) \):
\( f(3)=4(3)-(3)^{2}=12 - 9 = 3 \). So the point is \( (3, 3) \).

Step7: Find the equation of the tangent line

Use point - slope form \( y - y_{1}=m(x - x_{1}) \), where \( (x_{1},y_{1})=(3, 3) \) and \( m=-2 \):
\( y - 3=-2(x - 3) \). Simplify to slope - intercept form:
\( y - 3=-2x + 6\Rightarrow y=-2x + 9 \).

Part 1 (Derivative):

Answer:

The derivative of \( f(x)=4x - x^{2} \) is \( \boldsymbol{f'(x)=4 - 2x} \).

Part 2 (Tangent Line):