1. ## A couple of problems about derivitives

The directions state, a. Find the difference quotient of f.
b. f'(c) by comparing the limit of the difference quotient. Also, c is a constant.
h=change in x.
1.f(x)=f(x)=2-x^2 at c=0. the answer states: for a. -h and b. 0. For everytime I did the derivative formula, I get (2x-h). But when i plug 2 in for x and x^2 for h, it gets the right answer and that is what got me confused. Is it alwayz the case to use h when doing the derivative formula? If not, what are the exceptions?
2. Find the standard form equation for the tangent line to y=f(x)at the specified point.
f(x)= (x^2)-3x-5 where x=-2, the answer states:7x+y+9=0 I would know what rule to use if it had only an f(x) and a g(x). But on this one, I am not too sure.
3. Find the coordinates of each point on the graph of the given function where the tangent is horizontal.
f(x)=2x^3-7x^2+8x-3. the answer states: (1,0) and (4/3,-1/27). Same problem as #2 but not sure on the latter steps sense it is in coordinate form.

2. Originally Posted by driver327
The directions state, a. Find the difference quotient of f.
b. f'(c) by comparing the limit of the difference quotient. Also, c is a constant.
h=change in x.
1.f(x)=f(x)=2-x^2 at c=0. the answer states: for a. -h and b. 0. For everytime I did the derivative formula, I get (2x-h). But when i plug 2 in for x and x^2 for h, it gets the right answer and that is what got me confused. Is it alwayz the case to use h when doing the derivative formula? If not, what are the exceptions?
2. Find the standard form equation for the tangent line to y=f(x)at the specified point.
f(x)= (x^2)-3x-5 where x=-2, the answer states:7x+y+9=0 I would know what rule to use if it had only an f(x) and a g(x). But on this one, I am not too sure.
3. Find the coordinates of each point on the graph of the given function where the tangent is horizontal.
f(x)=2x^3-7x^2+8x-3. the answer states: (1,0) and (4/3,-1/27). Same problem as #2 but not sure on the latter steps sense it is in coordinate form.
Hello,

1. You have a function f with the equation y = f(x)
2. You have a point P(a, f(a)) which is placed on the graph of f.
3. Now the x-value changes a little bit. The new x-value is (a+h)
4. You get another point on the graph Q(a+h, (f(a+h))
5. Now you can calculate the slope between these 2 points:
$m = \frac{f(a+h) - f(a)}{(a+h)-a}$
6. This is the difference quotient for x = a.
7. Because you can use this equation for every x of the domain of f normally the equation is written like this:
$m = \frac{f(x+h) - f(x)}{(x+h)-x} = \frac{f(x+h) - f(x)}{h}$

8. $\lim_{h \rightarrow 0}(m) = f'(x)$

To #1:
$f(x) = 2-x^2$

$m = \frac{2-(x+h)^2 - (2-x^2)}{h} = \frac{2-x^2 - 2hx-h^2 -2 +x^2 }{h}=\frac{-2hx-h^2}{h}=-2x-h$ . Difference quotient

According to your problem x = 0 therefore m = -h

$f'(x) = \lim_{h \rightarrow 0}(-2x-h) = -2x$ . First derivation.

According to your problem x = 0 therefore f'(0) = 0

To #2:
$f(x) = x^2-3x-5$ then $f'(x) = 2x-3$

If x = -2 the f(-2) = 5 that means the point P(-2, 5) lies on the graph of f.
The gradient of the graph at x = -2 must be the same as the slope of the tangent in P.
$m_{tangent} = f'(-2) = 2(-2)-3 = -7$

Now use point-slope-formula of a straight line to get the equation of the tangent:
$(y-y_1) = m \cdot (x-x_1)$ Plug in all known values into this equation and solve for y:

$(y-5) = (-7) \cdot (x-(-2)) \Longleftrightarrow y = -7x-14+5 \Longleftrightarrow y = -7x-9$

To#3:
A horizontal line is a parallel to the x-axis that means it has the slope zero:
The slope of the tangent is the same as the gradient of the function:

$f(x) = 2x^3-7x^2+8x-3 \Longrightarrow f'(x) = 6x^2 - 14x + 8$

Now f'(x) = 0. Therefore:

$6x^2-14x+8 = 0 \Longleftrightarrow (3x-4)(2x-2)=0 \Longleftrightarrow x = \frac{4}{3} \vee x = 1$

Plug in these x-values into the equation of the function to get the second coordinate of the tangent points.