find a first degree polynomial function p(1) whose value and slope agree with the value and slope of f at x=c. i think you use taylor series for this
f(x)=4/sqrt(x), c=1
i'm totally confused on how to do it. i know you find the derivative but how do i get the function. this is number five on my homework so if you help me on this, i will be able to do the rest.
A first degree polynomial is a straight line of the form y = mx + b.
They are asking you for the equation of a line tangent to 4/sqrt(x) = 4 x^-(1/2) at x = 1
The value of y=f(x) at that point is y = 4
The value of the slope is
dy/dx = -2x^(-3/2), and when x=1, that slope is -2.
The equation of the tangent line is
(y-4)/(x-1) = -2
Convert that to y = mx + b form.
y-4 = -2x + 2
y = -2x + 6
To find a first degree polynomial function p(1) whose value and slope agree with the value and slope of f at x=c, follow these steps:
1. Start with the function f(x) and the given value of c. In this case, f(x) = 4/sqrt(x) and c = 1.
2. Find the derivative of f(x) to get the slope of the function. The derivative of 4/sqrt(x) is -2x^(-3/2).
3. Plug in c into the original function f(x) to get the value of f at x=c. In this case, f(1) = 4/sqrt(1) = 4.
4. Use the slope and value obtained from steps 2 and 3 to construct the equation of a tangent line to the function f at x=c. The equation will be in the form y = mx + b, where m represents the slope and (1, f(1)) represents a point on the line. The equation will be y = -2x + 6.
Therefore, the first degree polynomial function p(1) that satisfies the conditions is p(1) = -2(1) + 6 = 4.
You can use the Taylor series to approximate more complex functions, but in this case, a first degree polynomial is sufficient to match the value and slope at x=c.