Way up to x equals 7, including x equals 7. So it's defined for negativeġ is less than or equal to x. This function is not definedįor x is negative 9, negative 8, all the way down or all the way What is its domain? Well, exact similar argument. Is less than or equal to x, which is less thanĬondition right over here, the function is defined. So the domain of thisĭefined for any x that is greater than orĮqual to negative 6. Wherever you are, to find out what the value of It only starts getting definedĪt x equals negative 6. It's not defined for xĮquals negative 9 or x equals negative 8 and 1/2 or Is equal to negative 9? Well, we go up here. We say, well, what does f of x equal when x Is the entire function definition for f of x. Right over here, we could assume that this What is its domain? So the way it's graphed One more point (0,6) would give 6>3 which is a true statement, and shading should include this point. If point is (1,5) you can do the same thing, 5 > 5, but this would be right on the line, so the line would have to be dashed because this statement is not true either. If you try points such as (0,0) and substitute in for x and y, you get 0 > 3 which is a false statement, and if you did it right, shading would not go through this point. So lets say you have an equation y > 2x + 3 and you have graphed it and shaded. The has to do with the shading of the graph, if it is >, shading is above the line, and ). Without the "equal" part of the inequality, the line or curve does not count, so we draw it as a dashed line rather than a solid line Of 20.408 m, then h decreases again to zero, as expected.The "equal" part of the inequalities matches the line or curve of the function, so it would be solid just as if the inequality were not there. `t = -b/(2a) = -20/(2 xx (-4.9)) = 2.041 s `īy observing the function of h, we see that as t increases, h first increases to a maximum What is the maximum value of h? We use the formula for maximum (or minimum) of a quadratic function. It goes up to a certain height and then falls back down.) (This makes sense if you think about throwing a ball upwards. We can see from the function expression that it is a parabola with its vertex facing up. So we need to calculate when it is going to hit the ground. Also, we need to assume the projectile hits the ground and then stops - it does not go underground. Generally, negative values of time do not have any Have a look at the graph (which we draw anyway to check we are on the right track): So we can conclude the range is `(-oo,0]uu(oo,0)`. We have `f(-2) = 0/(-5) = 0.`īetween `x=-2` and `x=3`, `(x^2-9)` gets closer to `0`, so `f(x)` will go to `-oo` as it gets near `x=3`.įor `x>3`, when `x` is just bigger than `3`, the value of the bottom is just over `0`, so `f(x)` will be a very large positive number.įor very large `x`, the top is large, but the bottom will be much larger, so overall, the function value will be very small. As `x` increases value from `-2`, the top will also increase (out to infinity in both cases).ĭenominator: We break this up into four portions: To work out the range, we consider top and bottom of the fraction separately. So the domain for this case is `x >= -2, x != 3`, which we can write as `[-2,3)uu(3,oo)`. (Usually we have to avoid 0 on the bottom of a fraction, or negative values under the square root sign). In general, we determine the domain of each function by looking for those values of the independent variable (usually x) which we are allowed to use. For a more advanced discussion, see also How to draw y^2 = x − 2.
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