8x Show that the three functions: f₁(x) = ex-8, f₂(x) = e³x, f(x) = 4(2x-2) are linearly dependent.

Q: Find the domain of f(x) = Vx + 3 + Vx - 1.

A: Given that the function f(x) = x+3 +x-1 To find the domain of f(x).

Q: f(x) = Ve

A:

Q: Consider the function y = (a) Find the x- and y-intercepts of f(x)

A: An x-intercept of a function f(x) is a point on the x-axis at which the graph of the function f(x)…

Q: Find  f ′(a)  f(x) = 2x4,    a = −1

A:

Q: Determine whether the given set of functions is linearly independent on the interval (-o, ∞). f1(x)…

A: The solution is given as

Q: A function f(x) = v1 – x² is

A:

Q: Differentiate

A:

Q: Let f(x, y) = 2x - 4y + 7. Find the following. (a) f(3,-1) (b) f(-5, 1) (c) f(-5, -4) (d) f(0, 7)

A: Since you have posted multiple subparts, as per our policy we will answer first three subparts.…

Q: Let f be a linear function for which f(−4) = −9 and f(4) = 15. Determine f(0).   f(0) =

A:

Q: Describe the end behavior of each function. 7) f(x)= x -2x'+x-1 8) f(x)=-x² + 8x - 17

A:

Q: Find the linearization of f(x) =2x² + 4x – 3 at x= -1

A:

Q: Show that F(x) = 4x² + 6x +5 is anti-derivatives of f(x) = 8x + 6

A:

Q: b) h(x) - 13 — 2х - х?). c) h(x) = V3 - x 5, d) h(x) = Vx³ + 4x2 + 1.

A:

Q: Find y as a function of a if y(0) = 8, 3/(0) = 8, "(0) = 25, "(0) = 0. y(x) = || (4) 10y" +25y"=0, 1

A:

Q: Linearize the function f(x) = x3 + 2x + 4 at a = 3.

A: To linearize a function  at certain point x=a , first we need to evaluate the function value at a (…

Q: f(x)=xnex

A: f(x)=xn exThe doamin of the function is when n is even and odd is x∈ℝand the range of the function…

Q: A non-linear function is given by f(z) = a²-1 Where a is a constant. a) Obtain the linearized form…

A:

Q: Find the point on the graph of f(x)=4−x2 that is closest to the point (0,1).

A: To find the point on the graph of fx=4-x2 that is closest to the point (0,1)

Q: Find the derivative 3a+bt3 2b f(t) =+ V6at6. 5 3t

A:

Q: If f(x) = a, a > 0, then f'(12)

A: Please refer the attached image for complete solution. THANK YOU.