A differentiable function f is increasing on an interval if its derivative f'(x) is greater than or equal to 0 (strictly increasing if f'(x) > 0) for all x in that interval. It is decreasing if its derivative f'(x) is less than or equal to 0 (strictly decreasing if f'(x) < 0) for all x in that interval. So the sign of the first derivative tells us whether the function rises or falls.

f'(x) > 0 increasing; f'(x) < 0 decreasing

Find the derivative: f'(x) = 2x - 4 = 2(x - 2). Set f'(x) = 0 to get the critical point: x = 2. For x < 2, f'(x) < 0, so the function is decreasing on (-infinity, 2). For x > 2, f'(x) > 0, so the function is increasing on (2, infinity). Thus f is decreasing on (-infinity, 2) and increasing on (2, infinity).

The slope of the tangent to a curve y = f(x) at a point is the value of the derivative dy/dx at that point. The normal is the line perpendicular to the tangent at that point, so its slope is the negative reciprocal of the tangent's slope, that is, slope of normal = -1/(dy/dx). Using the point-slope form with these slopes gives the equations of the tangent and the normal.

slope of normal = -1/(slope of tangent)

The slope of the tangent is dy/dx = 2x. At the point (1, 1), the slope = 2(1) = 2. Using the point-slope form y - y1 = m(x - x1) with m = 2 and (x1, y1) = (1, 1): y - 1 = 2(x - 1), so y - 1 = 2x - 2, giving y = 2x - 1. This is the equation of the tangent.

y - y1 = m(x - x1)

Find the first derivative: f'(x) = 3x^2 - 3 = 3(x^2 - 1). Set f'(x) = 0: x^2 = 1, so x = 1 or x = -1. Second derivative: f''(x) = 6x. At x = 1, f''(1) = 6 > 0, so there is a local minimum; minimum value f(1) = 1 - 3 + 2 = 0. At x = -1, f''(-1) = -6 < 0, so there is a local maximum; maximum value f(-1) = -1 + 3 + 2 = 4.

f'(x) = 0; use f''(x) sign

Let f be a function and c be a point where f'(c) = 0 (a critical point). By the second derivative test: if f''(c) < 0, then f has a local maximum at c; if f''(c) > 0, then f has a local minimum at c; and if f''(c) = 0, the test fails and we must use the first derivative test (or higher derivatives) to decide.

Let the two numbers be x and 16 - x (since their sum is 16). Their product P = x(16 - x) = 16x - x^2. To maximise, differentiate: dP/dx = 16 - 2x, and set it to zero: 16 - 2x = 0, so x = 8. The second derivative d2P/dx2 = -2 < 0, confirming a maximum. So the numbers are x = 8 and 16 - 8 = 8, and the maximum product is 8 x 8 = 64.

Integration is the reverse (inverse) process of differentiation; it is used to find a function whose derivative is given. The indefinite integral of a function includes an arbitrary constant of integration C. The power rule for integration is: the integral of x^n dx = x^(n + 1)/(n + 1) + C, valid for n not equal to -1. For n = -1, the integral of (1/x) dx = ln|x| + C.

integral of x^n dx = x^(n+1)/(n+1) + C

Integrate term by term using the power rule. The integral of x^3 dx = x^4/4, the integral of 2x dx = x^2, and the integral of 1 dx = x. Adding a single constant of integration, the integral of (x^3 + 2x + 1) dx = x^4/4 + x^2 + x + C.

integral of x^n dx = x^(n+1)/(n+1) + C

The standard integrals are: the integral of sin x dx = -cos x + C; the integral of cos x dx = sin x + C; and the integral of e^x dx = e^x + C. Also, the integral of sec^2 x dx = tan x + C. These are obtained by reversing the corresponding derivative formulas.

integral of sin x dx = -cos x + C

Use the substitution t = 1 + x^2, so dt = 2x dx. The integral becomes the integral of (1/t) dt = ln|t| + C. Substituting back t = 1 + x^2, the answer is ln(1 + x^2) + C. (Note that 1 + x^2 is always positive, so the modulus sign is not needed.)

integral of f'(x)/f(x) dx = ln|f(x)| + C

Integration by parts uses the integral of u v dx = u (integral of v) - integral of (u' times integral of v). Take u = x (so u' = 1) and dv = e^x dx (so integral of dv = e^x). Then integral of x e^x dx = x e^x - integral of (1 x e^x) dx = x e^x - e^x + C = e^x(x - 1) + C.

integral u dv = u v - integral v du

Write 1/((x - 1)(x - 2)) = A/(x - 1) + B/(x - 2). Then 1 = A(x - 2) + B(x - 1). Putting x = 1: 1 = A(-1), so A = -1. Putting x = 2: 1 = B(1), so B = 1. Hence the integral = integral of [-1/(x - 1) + 1/(x - 2)] dx = -ln|x - 1| + ln|x - 2| + C = ln|(x - 2)/(x - 1)| + C.

(1) The integral from a to b of f(x) dx = - the integral from b to a of f(x) dx (interchanging the limits changes the sign). (2) The integral from a to b of f(x) dx = the integral from a to c of f(x) dx + the integral from c to b of f(x) dx (splitting the interval at a point c). (3) The integral from 0 to a of f(x) dx = the integral from 0 to a of f(a - x) dx. Also, for an odd function the integral from -a to a is 0, and for an even function it is twice the integral from 0 to a.

integral(a to b) f(x) dx = -integral(b to a) f(x) dx

First find the indefinite integral: the integral of x^2 dx = x^3/3. Now apply the limits using the fundamental theorem of calculus: the integral from 0 to 1 of x^2 dx = [x^3/3] evaluated from 0 to 1 = (1^3/3) - (0^3/3) = 1/3 - 0 = 1/3.

integral(a to b) f(x) dx = F(b) - F(a)

The integral of sin x dx = -cos x. Applying the limits from 0 to pi/2: the integral from 0 to pi/2 of sin x dx = [-cos x] from 0 to pi/2 = (-cos(pi/2)) - (-cos 0) = (-0) - (-1) = 0 + 1 = 1. So the value of the integral is 1.

integral of sin x dx = -cos x

Integrate term by term. The integral of (1/x) dx = ln|x| and the integral of e^x dx = e^x. Adding the constant of integration, the integral of (1/x + e^x) dx = ln|x| + e^x + C.

integral of (1/x) dx = ln|x| + C

The function f(x) = x^3 is an odd function, because f(-x) = (-x)^3 = -x^3 = -f(x). By the property of definite integrals, the integral from -a to a of an odd function is zero. Therefore the integral from -1 to 1 of x^3 dx = 0. (We can also verify: [x^4/4] from -1 to 1 = 1/4 - 1/4 = 0.)

integral(-a to a) of odd function = 0

The area bounded by the curve y = f(x), the x-axis, and the ordinates x = a and x = b (with f(x) >= 0) is given by the definite integral: Area = the integral from a to b of y dx = the integral from a to b of f(x) dx. The area between two curves y = f(x) (upper) and y = g(x) (lower) from x = a to x = b is Area = the integral from a to b of [f(x) - g(x)] dx.

Area = integral(a to b) f(x) dx

The area is given by the definite integral of y with respect to x from 0 to 3: Area = the integral from 0 to 3 of x^2 dx = [x^3/3] from 0 to 3 = (3^3/3) - (0) = 27/3 = 9. So the required area is 9 square units.

Area = integral(a to b) y dx