By radical, I am assuming that you mean square root, not cube root, quartic root, or otherwise. If this is the case, then we can use fractional exponents to help.
Change sqrt(x) to x^(1/2), or x to the one half power. Then we take a radical of a radical which becomes
sqrt(x^(1/2)) = (x^(1/2))^(1/2) = x^(1/4). When we raise a power to a power, we multiply exponents. So the answer to the square root of the square root of x is x to the one fourth power, or the 4th root of x.
If you set a function equal to zero and solve for x, then you are finding where the function crosses the x-axis.
We need more information. Is there a limit or integral? The theorem states that the deivitive of an integral of a function is the function
DomainFind the domain of y = x + 4Domain simply means "what numbers can I use for X that give me an answer for Y?"Domain is looking for the acceptable X values that work in the equation y = x +4When looking for the domain, the rule of thumb is:* No fraction? * No radical? (sqrt, cube root, etc) * No problems! If you don't see a fraction, or a radical, or both in your equation, the domain will always be "All real numbers". This means you can pick any number you want, plug it in for X, and you'll get an answer for Y.If you do happen to have a fraction, radical (or both)...----Think about what "breaks" a fraction. What makes a fraction not work?If you try to divide by 0, you get an error, or undefined on your calculator.for example:Find the domain of y = 1/(x -1)1/x is a fraction, and we know having a 0 on the bottom would make it not work.So what we do is say the domain is "all real numbers, except" and then find out what numbers break the function and fill that in later.We set the bottom of the fraction = 0 to find out exactly which numbers will break it, and then solve for X.x - 1= 0x = 1So we know that is we plug in a 1, it will break the fraction.The domain of y = 1/(x - 1) is "All real numbers, except 1"---- If you have a radical expression in your function:example. y = sqrt(x+2)We need to know what breaks a radical expression, what number(s) won't work?Well, sqrt(0) is OK. it = 0.sqrt(1) is ok, it = 1.But what happens if I try sqrt(-1). Try it on your calculator.You get an error message, right?This is because you cannot take the root of a negative number. (at least, not yet)This means, negative numbers break a function with a radical. So similarly to how we found the numbers that broke a fraction, we'll set what's inside the radical less than 0. Since negative numbers are less than 0.x + 2 < 0 and solve for xx < -2So any number less than -2 will break my radical.Your answer would be, "All real numbers >= -2" (since anything less than -2 is broken, but -2 is still OK)To check:y = sqrt( -2 + 2) = sqrt (0) = 0 - OKy = sqrt( -3 + 2) = sqrt(-1) - breaksI chose a number smaller than -2 to check.---- If you get a problem where they use both fractions AND radicals, just use both techniques.Ex: y = 1/sqrt(x + 2)We know that having a 0 on the bottom of a fraction breaks it, but it OK for a radical to have a 0 in it. We need to combine both rules together.Take the inside of the radical and set it less than(what breaks the radical) or equal to 0 (what breaks the fraction).x + 2
You take the derivative of the function, then solve the inequality:derivative > 0 for increasing, orderivative < 0 for decreasing.
A function--namely a parabola (concave up). To "evaluate" this function you would need an x value and would find the resulting y value. To "solve" this function, you would probably be given a y value and asked to find the corresponding x value(s).
Square both sides of the equation to get rid of the radical sign. Then just solve as you normally would. Good luck! :-)
This is a combination of two functions, where you apply the first function and get a result and then fill that answer into the second function. OR These are what you get when you take the output of one function and use it to solve the output of the next function.
This is a combination of two functions, where you apply the first function and get a result and then fill that answer into the second function. OR These are what you get when you take the output of one function and use it to solve the output of the next function.
It depends on what you mean by solve: simplify, evaluate or rationalise the denominator. The answer will also depend on the radical expression.
The property that is essential to solving radical equations is being able to do the opposite function to the radical and to the other side of the equation. This allows you to solve for the variable. For example, sqrt (x) = 125.11 [sqrt (x)]2 = (125.11)2 x = 15652.5121
First, get the radical by itself. Then, square both sides of the equation. Then just solve the rest.
the radical of 3 + the radical of 1/3
prime
to simplify the radicand
It really depends on the specific function, and what you want to solve for.
What square root property is essential to solve any radical equation involving square root?
That will depend on exactly how the equation is formed. In many cases, you can apply the inverse function to the outside first. Here is an example:sin(ln(x)) = ... To solve for "x", FIRST apply the inverse function of the sine (i.e., arcsin) to both sides of the equation. Next, apply the inverse of the natural logarithm to both sides. In this case, the exponential function (raise "e" to the power of the entire expression on both sides).