Puede tomar una gota mejor para enter
Please note that your x^3/4 is ambiguous. Did you mean (x^3) divided by 4
or did you mean x to the power (3/4)? I will assume you meant the first, not the second. Please use the "^" symbol to denote exponentiation.
If we have a function f(x) and its derivative f'(x), and a particular x value (c) at which to begin, then the linearization of the function f(x) is
f(x) approx. equal to [f '(c)]x + f(c)].
Here a = c = 81.
Thus, the linearization of the given function at a = c = 81 is
f(x) (approx. equal to) 3(81^2)/4 + [81^3]/4
Note that f '(c) is the slope of the line and is equal to (3/4)(81^2), and f(c) is the function value at x=c, or (81^3)/4.
What is the linearization of f(x) = (x^3)/4, if c = a = 81?
It will be f(x) (approx. equal to)
Answer:A. -1.11x^2+ 3.74+ 10.64
Step-by-step explanation:
a. p. e. x. (just took the test)
(-2,2)(2,-8)
slope = (-8 - 2) / (2 - (-2) = -10/4 = -5/2
y = mx + b
slope(m) = -5/2
(-2,2)...x = - 2 and y = 2
now we sub and find b, the y int
2 = -5/2(-2) + b
2 = 5 + b
2 - 5 = b
-3 = b
so ur equation is : y = -5/2x - 3
Answer:
(5,-2) or y=3(x-5)^2 -2
Step-by-step explanation:
