MATH 133 -- Intensive Calculus for Science 1

Lab Day 1 Solutions

1.

> f:=x->sin(x)-x*cos(2*x);

> plot(f(x),x=-4..4);

2.

> plot(D(f)(x),x=-4..4);

3. By looking at the graph y = f '(x) at x = 1, we see f '(1) = 2.8 (approx.)

The equation of the tangent line is approximately:

y - 1 = 2.8( x- 1) or y = 2.8 x - 1

> tang:=x->2.8*x-1;

> plot({f(x),tang(x)},x=-4..4);

Of course, this is not exactly tangent since we estimated the slope and the

y- coordinate of the point.

4. a f '(x) is positive roughly -4 <= x <= -3.3, -1.7 <= x <= 1.7, and 3.3 <= x <= 4.

b. On each of these intervals, f is increasing. In general, if the derivative of f is positive

on an interval, then f is increasing on that interval because the tangent lines slope

up to the right.

c. f '(x) is negative roughly -3.3 <= x <=-1.7 and 1.7 <= x <= 3.3.

d. On each of these intervals, f is decreasing. In general, if the derivative of f is negative

on an interval, then f is decreasing on that interval because the tangent lines slope

down to the right.

e. f '(x) = 0 roughly at x = -3.3, -1.7, 0, 1.7, 3.3. At each of those points, the

tangent line to y = f (x) is horizontal. (Note this includes x = 0, which is not

a maximum or a minimum.

5.

> plot((D@@2)(f)(x),x=-4..4);

6. a f ''(x) is positive roughly -2.6 <= x <= -1.1, 0 <= x <= 1.1, and 2.6 <= x <= 4.

b. On each of these intevals, f ' is increasing. In general, if the second derivative of f

is positive on an interval, then f ' is increasing on that interval because f '' is

the derivative of f '. On those intervals, f is concave up.

c. f ''(x) is negative roughly -4 <= x <=-2.6, -1.1 <= x <= 0, and 1.1 <= x <= 2.6.

d. On each of these intervals, f ' is decreasing. In general, if the second derivative of f

is negative on an interval, then f ' is decreasing on that interval because f '' is

the derivative of f '. On those intervals, f is concave down.