Math  /  Algebra

Question1) Determina dominio intersezioni con gli assi e segno della funzione y=1x310x2+25xy=\frac{1}{x^{3}-10 x^{2}+25 x} Indica se la funzione è pari, dispari o né pari né dispari. Disegna le parti di piano in cui la funzione esiste

Studdy Solution

STEP 1

1. The function is given by y=1x310x2+25x y = \frac{1}{x^3 - 10x^2 + 25x} .
2. The domain of the function is determined by the values of x x that do not make the denominator zero.
3. Intersections with the axes occur when x=0 x = 0 (y-intercept) and y=0 y = 0 (x-intercepts).
4. The sign of the function is determined by the sign of the denominator.
5. A function is even if f(x)=f(x) f(-x) = f(x) and odd if f(x)=f(x) f(-x) = -f(x) .

STEP 2

1. Determine the domain of the function.
2. Find the intersections with the axes.
3. Determine the sign of the function.
4. Check if the function is even, odd, or neither.
5. Sketch the regions of the plane where the function exists.

STEP 3

To find the domain, we need to determine where the denominator is non-zero:
x310x2+25x0 x^3 - 10x^2 + 25x \neq 0
Factor the denominator:
x(x210x+25)=0 x(x^2 - 10x + 25) = 0
x(x5)2=0 x(x - 5)^2 = 0
The solutions to this equation are x=0 x = 0 and x=5 x = 5 .

STEP 4

Find the intersections with the axes:
- **Y-intercept**: Set x=0 x = 0 in the function. Since the denominator becomes zero, there is no y-intercept.
- **X-intercepts**: Set y=0 y = 0 . Since the numerator is a constant (1), the function never equals zero, so there are no x-intercepts.

STEP 5

Determine the sign of the function by analyzing the sign of the denominator:
The critical points are x=0 x = 0 and x=5 x = 5 . Analyze intervals:
- (,0) (-\infty, 0) : Choose x=1 x = -1 , denominator is negative. - (0,5) (0, 5) : Choose x=1 x = 1 , denominator is positive. - (5,) (5, \infty) : Choose x=6 x = 6 , denominator is positive.
The function is negative on (,0) (-\infty, 0) and positive on (0,5) (0, 5) and (5,) (5, \infty) .

STEP 6

Check if the function is even, odd, or neither:
Calculate f(x) f(-x) :
f(x)=1(x)310(x)2+25(x)=1x310x225x f(-x) = \frac{1}{(-x)^3 - 10(-x)^2 + 25(-x)} = \frac{1}{-x^3 - 10x^2 - 25x}
Compare f(x) f(-x) and f(x) -f(x) :
Since f(x)f(x) f(-x) \neq f(x) and f(x)f(x) f(-x) \neq -f(x) , the function is neither even nor odd.

STEP 7

Sketch the regions of the plane where the function exists:
The function exists in the intervals (,0) (-\infty, 0) , (0,5) (0, 5) , and (5,) (5, \infty) .
- In (,0) (-\infty, 0) , the function is negative. - In (0,5) (0, 5) , the function is positive. - In (5,) (5, \infty) , the function is positive.
The domain of the function is (,0)(0,5)(5,) (-\infty, 0) \cup (0, 5) \cup (5, \infty) , and the function is neither even nor odd.

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