suppose function f,g,h and l are defined as follow:f(x)=4xsquared-5x+1,g(x)=-1/2x-3,h(x)=squared root(2-x)squared+1/2x,l(x)=square root2-x-3.
.write down Df and solve inequality f(x)<0
.write down Dg and solve 4g(x)=8
.write down Dh,Dg+h and solve the (g+h)(x)=0
.write down Dg.l without first calculating(g.l)(x)
.write down Dl/g without first calculating l/g(x)
i need help pls
Sure, I can help you with that. Let's break it down step by step.
1. Df and solving the inequality f(x) < 0:
Df represents the set of all real numbers for which the function f(x) is defined. In this case, f(x) is a quadratic function, which means it is defined for all real numbers.
To solve the inequality f(x) < 0, we start by finding the roots of the quadratic equation f(x) = 0. The roots can be found using the quadratic formula or factoring.
f(x) = 4x^2 - 5x + 1
To find the roots, we set f(x) = 0:
0 = 4x^2 - 5x + 1
Factoring the quadratic equation, we have:
0 = (4x - 1)(x - 1)
Setting each factor equal to 0, we get:
4x - 1 = 0 --> 4x = 1 --> x = 1/4
x - 1 = 0 --> x = 1
So the roots are x = 1/4 and x = 1.
Now, to solve the inequality f(x) < 0, we need to find the intervals where f(x) is negative. We can use a sign chart or test points:
Let's test the interval (1/4, 1/4 + δ):
f(0) = 4(0)^2 - 5(0) + 1 = 1 (positive)
f(1/2) = 4(1/2)^2 - 5(1/2) + 1 = -1/2 (negative)
Since the sign changes from positive to negative between 0 and 1/2, the solution to the inequality f(x) < 0 is:
x ∈ (0, 1/2)
Therefore, Df is the set of all real numbers, and the solution to f(x) < 0 is x ∈ (0, 1/2).
2. Dg and solving 4g(x) = 8:
Dg represents the set of all real numbers for which the function g(x) is defined. In this case, g(x) is defined for all real numbers except x = 0 (since division by 0 is undefined).
To solve 4g(x) = 8, we can first divide both sides of the equation by 4:
g(x) = 8/4
g(x) = 2
So, g(x) = 2 for all values of x except x = 0. Therefore, Dg is the set of all real numbers excluding x = 0.
3. Dh, Dg+h, and solving (g+h)(x) = 0:
Dh represents the set of numbers for which the function h(x) is defined. In this case, h(x) is defined for all real numbers.
To find Dg+h, we need to find the set of numbers for which both g(x) and h(x) are defined. Since both g(x) and h(x) are defined for all real numbers, Dg+h is also all real numbers.
To solve (g+h)(x) = 0, we substitute the given functions:
(g+h)(x) = g(x) + h(x) = 0
Since g(x) = 2 for all x, we can rewrite the equation as:
2 + h(x) = 0
Subtracting 2 from both sides, we get:
h(x) = -2
So, the solution to (g+h)(x) = 0 is h(x) = -2 for all values of x.
4. Dg.l without first calculating (g.l)(x):
To find Dg.l, we need to find the set of numbers for which both g(x) and l(x) are defined.
From the given functions:
g(x) = -1/2x - 3
l(x) = sqrt(2 - x) - 3
Both g(x) and l(x) are defined for all real numbers, except when the square root of (2 - x) is not defined (i.e., when 2 - x < 0).
To find the domain of l(x), we set 2 - x ≥ 0:
2 - x ≥ 0
x ≤ 2
Therefore, Dg.l is the set of all real numbers where x ≤ 2.
5. Dl/g without first calculating l/g(x):
To find Dl/g, we need to find the set of numbers for which both l(x) and g(x) are defined.
From the given functions:
l(x) = sqrt(2 - x) - 3
g(x) = -1/2x - 3
Both l(x) and g(x) are defined for all real numbers, except when the square root of (2 - x) is not defined (i.e., when 2 - x < 0) and when x = -4 (since division by 0 is undefined).
To find the domain of l(x) / g(x), we consider the conditions:
1. 2 - x ≥ 0,
2. x ≠ -4
Therefore, Dl/g is the set of all real numbers where 2 - x ≥ 0, and x ≠ -4. In interval notation, it can be written as (-∞, -4) ∪ (-4, ∞).
Sure! Let's go step by step and solve each part of the question.
1. Df represents the domain of the function f(x). To find Df, we need to determine the values of x for which the function is defined. In this case, f(x) is a polynomial, so it is defined for all real values of x. Therefore, Df is the set of all real numbers, or Df = (-∞, +∞).
To solve the inequality f(x) < 0, we can use various methods, such as factoring, the quadratic formula, or graphing. Here, we have a quadratic function f(x) = 4x^2 - 5x + 1. We can find the x-intercepts by setting f(x) = 0 and solving for x.
To solve, 4x^2 - 5x + 1 = 0, we can factor or use the quadratic formula. However, for this example, let's use factoring. Find two numbers that multiply to give the product of 4 and 1 (the coefficient of x^2 and the constant term), and also add up to give -5 (the coefficient of x). In this case, the numbers are -4 and -1:
4x^2 - 5x + 1 = (4x - 1)(x - 1) = 0
Setting each factor equal to zero, we get:
4x - 1 = 0 --> 4x = 1 --> x = 1/4
x - 1 = 0 --> x = 1
Thus, the x-intercepts are x = 1/4 and x = 1. To determine where f(x) < 0, we can use a sign chart or test points. Testing points within each interval (before 1/4, between 1/4 and 1, and after 1) shows that the function is negative between 1/4 and 1. Therefore, the solution to the inequality f(x) < 0 is x ∈ (1/4, 1).
2. Dg represents the domain of the function g(x). Here, g(x) is a rational function involving division by x, so it is not defined for x = 0. Therefore, Dg = (-∞, 0) U (0, +∞), or all real numbers except x = 0.
To solve the equation 4g(x) = 8, we substitute g(x) = -1/2x - 3 into the equation:
4(-1/2x - 3) = 8
Simplifying:
-2x - 12 = 8
-2x = 20
x = -10
The solution to the equation 4g(x) = 8 is x = -10.
3. Dh represents the domain of the function h(x). In the expression for h(x), we have a square root. For the square root to be defined, the expression inside the square root must be greater than or equal to zero. Therefore, 2 - x ≥ 0, which means x ≤ 2. Also, the denominator of 1/2x must be non-zero, so x ≠ 0. Thus, Dh = (-∞, 0) U (0, 2].
To solve (g + h)(x) = 0, we substitute g(x) = -1/2x - 3 and h(x) = √(2 - x)^2 + 1/2x into the equation:
(-1/2x - 3) + (√(2 - x)^2 + 1/2x) = 0
Simplifying:
-1/2x - 3 + √(2 - x)^2 + 1/2x = 0
The equation is quite complicated, but we can simplify it further:
-3 + √(2 - x)^2 = 0
Since the square root of a number is always non-negative, the expression √(2 - x)^2 is always non-negative. Therefore, the equation -3 + √(2 - x)^2 = 0 has no real solutions. This means that there are no values of x that satisfy the equation (g + h)(x) = 0.
4. Dg⋅l represents the domain of the function g(x) multiplied by l(x). Since multiplying two functions means that both functions need to be defined at the given point to obtain the product, we need to find the intersection of their individual domains.
Dg = (-∞, 0) U (0, +∞) (from previous calculation)
Dl = {x | 2 - x ≥ 0} = (-∞, 2]
Therefore, the intersection of Dg and Dl is Dg⋅l = (-∞, 0) U (0, 2].
5. Dl/g represents the domain of the function l(x) divided by g(x). When dividing two functions, the denominator must not be equal to zero to avoid division by zero errors.
Dg = (-∞, 0) U (0, +∞) (from previous calculation)
To find the values of x where g(x) = 0, we set g(x) = -1/2x - 3 equal to zero:
-1/2x - 3 = 0
-1/2x = 3
x = -6
Thus, the value x = -6 makes g(x) equal to zero, which means we cannot divide by g(x) at x = -6. Therefore, the domain Dl/g is the set of all real numbers except x = -6.
I hope this helps! Let me know if you have any further questions.