If x is zero, then y12x is also zero. The answer is therefore 88.
let f be a function and f' the first derivative.If f'>0 the function is genuinely ascending.If f'
The general equation for a linear approximation is f(x) ≈ f(x0) + f'(x0)(x-x0) where f(x0) is the value of the function at x0 and f'(x0) is the derivative at x0. This describes a tangent line used to approximate the function. In higher order functions, the same concept can be applied. f(x,y) ≈ f(x0,y0) + fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0) where f(x0,y0) is the value of the function at (x0,y0), fx(x0,y0) is the partial derivative with respect to x at (x0,y0), and fy(x0,y0) is the partial derivative with respect to y at (x0,y0). This describes a tangent plane used to approximate a surface.
The answer depends on the nature of the function that defines the curve whose slope you want. If the function f(x) is differentiable, its slope is f'(x) = df(x)/dx and the value of the slope at a point when x = x0 is f'(x0), obtained by substituting x0 for x in f'(x).
When your input variable causes your denominator to equal zero. * * * * * A rational function of a variable, x is of the form f(x)/g(x), the ratio of two functions of x. Suppose g(x) has a zero at x = x0. That is, g(x0) = 0. If f(x0) is not also equal to 0 then at x = x0 the rational function would involve division by 0. But division by 0 is not defined. Depending on whether the signs of f(x) and g(x) are the same or different, as x approaches x0 the ratio become increasingly large, or small. These "infinitely" large or small values are the asymptotes of the rational function at x = x0. If f(x0) = 0, you may or may not have an asymptote - depending on the first derivatives of the two functions.
A tangent line is NEVER vertical to a function. It is vertical to the normal to the function - which is as far from vertical as you can get!The graph of a function, f(x) can have a tangent at a point. Let's call the point (x0,f(x0)). If f'(x) goes to positive infinity or f'(x) goes to negative infinity as x approaches x0 then f(x) has a vertical tangent at that point.
shut up and do your hw
For a quadratic function, there is one minimum/maximum (the proof requires calculus) and also it is either always convex or concave (prove is also calculus) it is continuous every where, hence, it can have a maximum of 2 roots. Graph it. If there is more than 2 roots, by Intermediate Value Theorem, it cannot be convex/concave everywhere. It will HAVE to have two intervals of increasing or decreasing. It can be easily proven that given any quadratic function f(x), if x = x0 is a minimum/maximum, and x=a != x0 is a root, then 2x0-a is also a root. It is still true that a = x0 as 2x0-x0=x0 implying it is the only root. But the concept of min/max requires Calculus to prove existence. So, this is Calculus, not algebra.
It's a method used in Numerical Analysis to find increasingly more accurate solutions to the roots of an equation. x1 = x0 - f(x0)/f'(x0) where f'(x0) is the derivative of f(x0)
so 10 x0 equals 0 people its times and x is diffrent to plus.
A function f(x), of a variable x, is said to have a limiting value of f(xo) as x approaches x0 if, given any value of epsilon, however small, it is possible to find a value delta such that |f(x) - f(x0)| < epsilon for all x such that |x - x0| < delta.The second inequality can be one-sided.
The answer is -13 1/3ohere is the detailed calculation for the problem:Let x0 be the angle, then;(180 - x0) - 2[180 - (90 - x0)] =40(180 -x0) - 2[90+x0]=40180 -x0 - 180 - 2x0=40-3x0=40hencex0= -13 1/3oAny comments are welcome
0! You said x0! anything x0=0!