Instruction

1

Follow finding the maximum value of the function, which in the interval has a finite number of critical points. To do this, calculate its

**value**at all points and at the ends of the segment. From the resulting numbers, choose the highest. The method of finding the greatest value**of the expression**is used to solve various application tasks.2

To do this, run the following steps: translate the task into the language of the function, select x, and through him to Express the required rate as a function of f(x). Using Analytics, find the largest and smallest values of the function at a certain interval.

3

Use the following examples to find the value of the function. Find the values of the function y=5-square root (4 – x2). Using the definition of the square root, we get 4 - x2 > 0. Solve the quadratic inequality, we get that -2 < x < 2. Divide the resulting interval into two, you get two inequalities -2 < x < 0 and 0 < x < 2.

4

Erect in the square each of the inequalities, then multiply all three parts by -1, add to them 4. Then enter the auxiliary variable and make the assumption that t = 4 - x2, where 0 < t < 4. A function of y, is the square root of the variable t on the interval is increasing and continuous. Therefore, the greatest

**value**of the function happens at the end of the period.5

Perform the inverse change of variables, as a result, you will get the following inequalities: 0 < root of (4 – x2) < 2. Add to all portions 5, before multiplying by -1, you get 3 < 5 - root of (4 – x2). < 5. Thus, the set of values of the function y = 5 - square root (4 – x2) is the interval [3; 5], and the largest

**value**, respectively, 5.6

Use the method of application properties of continuous functions to determine the highest

**value****of the expression**. In this case, use the numeric values taken by the expression at the specified interval. Among them, there is always a least**value**m and maximum**value**M. Between these numbers is a set of values of the function.# Advice 2: How to find the smallest root

For the solution of the quadratic equation and find the smallest root discriminant is calculated. The discriminant is equal to zero only if the polynomial has multiple roots.

You will need

- mathematical Handbook;
- calculator.

Instruction

1

Bring the polynomial to a square equation of the form ax2 + bx + c = 0 where a, b and c are arbitrary real numbers, with a in no case must not equal 0.

2

Substitute the values of the resulting quadratic equation into the formula for calculating the discriminant. This formula is as follows: D = b2 - 4ac. In that case, if D is greater than zero, the quadratic equation will have two roots. If D equals zero, both the calculated root, will be not only real, but equal. Third option: if D is less than zero, the roots will be a complex number. Calculate the value of the roots: x1 = (-b + sqrt (D)) / 2a and x2 = (-b - sqrt (D)) / 2a.

3

To calculate roots of a quadratic equation can also use the following formula: x1 = (-b + sqrt (b2 - 4ac)) / 2a and x2 = (-b - sqrt (b2 - 4ac)) / 2a.

4

Compare the two calculated root: root with the smallest value is the desired value you.

5

Not knowing the roots of a quadratic trinomial, you can easily find their sum and product. Use the vieta theorem, according to which the sum of the square roots of the trinomial represented in the form x2 + px + q = 0, is equal to the second ratio, that is, p, but with opposite sign. The product of the roots corresponds to the value of the free term q. In other words, x1 + x2 = – p, and x1x2 = q. For example, given the following quadratic equation: x2 – 5x + 6 = 0. To start, lay 6, two multiplier, and so that the sum of these multipliers was equal to 5. If you picked up the values correctly, then x1 = 2, x2 = 3. Test yourself: 3x2=6, 3+2=5 (as required, 5 with the opposite sign, i.e. "plus").

Note

Be careful not to make a mistake, placing signs!

Useful advice

The number with the sign "minus" is always less than positive. If I compare two negative values, then less of them will be the fact that the module is more.