Find the domain and range of f(x)= -2 + the square root of 16 - x^2?
16 - x^2>=0 (otherwise the square root is not a real number) x^2<=16 |x|<=4 -4<=x<=4 Domain=D=[-4;4] It's obvious that 0<=16 - x^2<=16 in D ==> 0<=sqrt(16 - x^2)<=4 in D ==> -2<=f(x)<=2 in D Range=[-2;2] To graph this function you need to make some transformations. First let y=f(x) ==> y = -2 + sqrt(16 - x^2) y+2=sqrt(16 - x^2) (y+2)^2=[sqrt(16 - x^2)]^2 (y+2)^2=16 - x^2 x^2+(y+2)^2=16 x^2+(y+2)^2=4^2 It's an equation of a circle with center at C(0;-2) and radius r=4. But the graph of your function is only the upper half of that circle, because Range=[-2;2] You may see the graphs of your functions at http://gcalc.net/
F(x)=square root 64-x^2. what are domain and range?
Domain is the set of values which you can put into the function. F(x) is a square root function which will not be defined for negative values inside the root so you need to solve 64 - x^2 >= 0 --> (8-x)(8+x) >=0 --> -8 <= x <= 8 --> [-8, 8] Range is the set of outputs from the function. From our domain F(x) is at its maximum when x = 0 (F(0) = sqr(64 - 0^2) = 8) and F(x) is at its minimum when x = -8 or x = 8 (F(8) = 0) So the range is [0,8]
Transformation of F(x)=square root of x?
I am supposed to write an equation for the transformation of f(x)=square root of x for this graph Hopefully this picture works: http://www.webassign.net/waplots/d/8/edbd1eed120afa9db62520879b7e24.gif I got the right answer, which is y=-square root of (-x+4)-5. With -x+4 being under the square root sign and -5 being outside. I'm just confused as to why it is +4 and not -4, since I moved the graph to the right.
Why can't the square root of 4 be -2 instead of 2, if -2 times -2 also equals 4?
Every positive real number has two square roots; one is positive and the other is negative. In addition, a square root of a number is one of its two equal factors, for example, 2 is a square root of 4 because (2)(2) = 4, but ‒2 is also a square root of 4 because (‒2)(‒2) = 4. In general, a real number “a” is a square root of a nonnegative number “b” if a² = b. By definition, the symbol √ b is used to designate the nonnegative or principal square root of any nonnegative real number b, e.g., √4 = 2 is used to indicate the nonnegative or principal square root of 4, and ‒√4 = ‒2 is used to indicate the negative square root of 4. Although the principal square root of a positive number is only one of its two square roots, the designation “the square root” is often used to refer to the principal or nonnegative square root √ b. This is why √4 = 2, rather than ‒2.
What is the inverse function of -1/7 square root 16 - x^2?
Do you mean: y = (-1/7) * sqrt(16 - x^2) -7y = sqrt(16 - x^2) 49y^2 = 16 - x^2 x^2 = 16 - 49y^2 x = +/- sqrt(16 - 49y^2) The inverse is: y = +/- sqrt(16 - 49x^2) Assuming we're dealing with real numbers only, note that in the original equation, y <= 0, because sqrt(16 - x^2) >= 0, so in the inverse equation, x <= 0 Note that the inverse is NOT a function!
How would one go about graphing [math]f(x) = x^3 - 8x^2 + 16x - 3[/math]?
Follow the below points to sketch the curve.Find the domain of the function, as it is a polynomial defined everywhere on the real line.Find the x and y intercepts and plot the points on the xy-axis.Find the critical points and inflection points by setting first derivative and second derivatives equal to zero and solve the equations.First , split the domain that is the real line into intervals based on inflection points.Take a sample point in the intervals and evaluate the second derivative at the points. If it is positive, the graph of the function some thing look like a concave up graph. Negative, it is concave down. Keep the point in your mind. Now, go to next point.Split the domain into intervals based on critical points and x-intercepts. Take a sample point in a interval and evaluate the value of the first derivative at the point. If the derivative is positive, the graph of the function is increasing in the interval . If it is negative,the graph of the function is decreasing. Now, draw the behavior of the function on each interval that you spitted down and merge all . You will get the graph of the required function.Hope this will help you :)
Find linear approximation of f(x)=sqrt(16-x) at a=0 and use it to approximate sqrt(15.9) and sqrt(16.1)?
First find the equation of the tangent line for f at x = 0: Point: (0, 4) Slope: f'(x) = -(1/2) * (16 - x)^(-1/2). At x = 0, m = (-1/2) * 16^(-1/2) = -1/8. So, the tangent line is given by y - 4 = (-1/8)(x - 0) ==> y = -x/8 + 4. For x near 0, we have f(x) = sqrt(16 - x) =(approx.) -x/8 + 4. So, sqrt(15.9) = sqrt(16 - 0.1) =(approx.) -0.1 / 8 + 4 = 3.9875 and sqrt(16.1) = sqrt(16 - (-0.1)) =(approx.) -(-0.1) / 8 + 4 = 4.0125. I hope this helps!