can help graph many functions. The first derivative of a function
is the slope of the tangent line for any point on the function!
Therefore, it tells when the function is increasing, decreasing or
where it has a horizontal tangent! Consider the following graph:
on the left side, the function is increasing and the slope of the
tangent line is positive. At the vertex point of the parabola,
the tangent is a horizontal
line, meaning f '(x) = 0 and on the right side the graph is decreasing
the slope of the tangent line is negative!
observations lead to a generalization for any function f(x) that has a
derivative on an interval I:
1) If f '(x) > 0 on an interval I, then the graph of f(x)
rises as x increases.
2) If f '(x) < 0 on an interval I, then the graph of f(x)
falls as x increases.
3) If f '(c) = 0, then the graph of f(x) has a horizontal tangent
at x = c. The function
may have a local maximum or minimum value, or a point of inflection.
are some graphs of each of the observations made above!
Some observations on
1) To be a minimum point,
the graph must change direction from decreasing to increasing.
2) To be a maximum
point, the graph must change direction from increasing to decreasing.
3) To be an inflection
point, the graph doesn't change direction. In the above example (
one in middle) it is increasing before the f '(c) = 0 and it is still
increasing after. You can also have one with the graph decreasing
on both sides.
Find the critical points(maximum, minimum or inflection points) of the
function f(x) = x3 + 3x2 - 4. Then graph the function.
Find the critical points by finding f '(x).
= 3x2 + 6x
the zeros by solving f '(x) = 0
3x2 + 6x = 0
+ 2) = 0
we have critical points at x = 0, and x = -2
these values into the original function to find the y values of the critical points. The
points are (0, -4) and (-2, 0)
Use the derivative to find where the graph is increasing and decreasing
by taking x values in each of the three areas formed by the two
critical points. The chart below shows the results
x < -2
x = -2
-2 < x < 0
x = 0
x > 0
chart shows that (-2, 0) is a local Max. and (0, -4) is a local
min. You can
tell because of the sign changes!
Find the zeros of the original function. These are the
x-intercepts. You can use synthetic division and factoring to
find the zeros! They are (-2, 0) (double root) and (1, 0)
Find the y-intercept. This is the constant of the original
function. (0, -4)
Now take the limit as x goes to both infinities of the original
Now put all these together and graph the function!
Find the critical points and graph f(x) = x4 - 8x2 + 9
Find the critical points by finding f '(x)
= 4x3 - 16x
f '(x) = 0 and solve
4x3 - 16x = 0 4x(x2 - 4) = 0 4x(x - 2)(x + 2) = 0
the critical points are at x = 0, x = 2 and x = -2 Substitute these into
the original function and the points are (0, 9), (2, -7), (-2, -7)
Use the derivative to find the slope in each of the four areas divided
by the three critical points. The chart is below
x < -2
x = -2
-2 < x < 0
x = 0
0 < x < 2
x = 2
x > 2
chart shows that (-2, -7) and (2, 7) are minimums and (0, 9) is a
maximum. Check out the signs on the chart to realize this!
Find the zeros of the original function by using your graphing
calculator. The zeros are not rational but irrational.
The y intercept is (0, 9)
Find the limits on infinity.
Put these together and graph the function!
The next section is one you are going to love! Yes,
your favorite and mine. Word problems related to maxima and
minima!! Can't wait!!