Derivative and tangent line

Let f(x)=4x-x^2. In this problem, we’re going to compute f'(x) using the difference quotient and use that to find the equation of a line that’s tangent to the graph of f.

1. Write down the down the difference quotient for f.
2. Find f'(x) by computing \lim_{h\to0} of your difference quotient.
3. Use f' to find the slope of the line that’s tangent to the graph of f at the point x=1.
4. Write down an equation of the line that’s tangent to the graph of f at the point x=1.

1. Different quotient:

2. f'(x):

3. Slope of the line that is tangent to the graph of f at the point x=1:

4. Equation of the line that is tangent to the graph of f at the point x=1:

@cmodlin2 and @rserra

I think you’ve both got a decent difference quotient written down but neither of you have really applied it by simplifying and then taking a limit.

I guess I don’t know where the integration symbol is coming from or why you would write it.? For that matter, we haven’t learned integration at all.

I kinda like @rserra’s answers to 3 and 4.

1. Difference Quotient : ([4(x+h)-(x+h)^2]-(4x-x^2))/h

2. f'(x) = \lim_{h\to0} \frac{(4(x+h)-(x+h)^2) - (4x - x^2)}{h} \\ = \lim_{h\to0} \frac{4x+4h-x^2-2xh-h^2 - 4x + x^2}{h} \\ = \lim_{h\to0} \frac{4h-2xh-h^2}{h} = \lim_{h\to0} \frac{h(4-2x-h)}{h} \\ = \lim_{h\to0}(4-2x-h) = 4-2x.

3. Slope of the line that is tangent to the graph of f at point x=1:

4. The Equation of the line that is tangent to the graph of f at the point x=1:

@mbradle7 This excellent - thanks! I did edit your part two to fall more in line with traditional LaTeX typesetting. In particular, I used $$stuff$$ to typeset the stuff and I used \\ to break lines. Remember that you can cntrl-click on the math to see the details.

Where does the “-3” on question 4 come from?

Difference quotient
\frac{[4(x+h)-(x+h)^2]-(4x-x^2)}{h}
2.
\lim_{h\to0}\frac{[4(x+h)-(x+h)^2]-(4x-x^2)}{h}
\lim_{h\to0}\frac{4x+4h-x^2-2xh-h^2-4x+x^2}{h}
\lim_{h\to0}\frac{4h-2xh-h^2}{h}
\lim_{h\to0}\frac{h(4-2x-h)}{h}
\lim_{h\to0}4-2x-h
(y\prime)=-2x+4
3.
Slope
y=-2m+4
-4=-2m
m=2
4
Equation
f(x)=4(1)-(1)^2
f(x)=3
y-3=2(x-1)
f(x)=2x+1

Well, the point-slope formula for a line states:

In particular, the y_0 represents the y-coordinate of the point (x_0,y_0) on the line. We’re given the x-coordinate, namely x_0=1 and we get the y-coordinate by plugging that into f. Thus,

The the y-y_0 on the left side is y-3.

1. f(x)=\frac{\left(\left(4\left(x+h\right)-\left(x+h\right)^2\right)-\left(4x-x^2\right)\right)}{h}
2. f’(x)= 4-2x
3. f’(1)= m=2
4. y-3=2(x-1)

The Difference Quotient:((4(x+h)-(x+h)^2)-4x+x^2)/h

2. f'(x) = \lim_{h\to0} \frac{(4(x+h)-(x+h)^2) - (4x - x^2)}{h} \\ = \lim_{h\to0} \frac{4x+4h-x^2-2xh-h^2 - 4x + x^2}{h} \\ = \lim_{h\to0} \frac{4h-2xh-h^2}{h} = \lim_{h\to0} \frac{h(4-2x-h)}{h} \\ = \lim_{h\to0}(4-2x-h) = 4-2x.
3. The slope of the line tangent to the point x = 1 on the graph is:

4.The equation of the tangent line of the graph of f at x = 1 is:

1.The difference quotient for f is:

2. f’(x) =

3. m=4-2(1) = 2
4. y-y_0=m(x-x_0) \\ =y-3=2(x-1) \\ = y=2(x-1)+3