(1+h): - 1 A. lim
h>0
h
f(0) =
х
a =
1
B. lim
h>0
cos(1 + h) +1
h
f(3) =
a =
7T
5(4+ h)" - 5120
C. lim
h->0
h
f(x) =
a =
4
o
D. lim
h 0
e4h - 1
h
f(x) =
a =
e
Х
Pls take a look at the picture

Each of these limits correspond to the derivative of some function f(x) at a point x = a.

Recall the limit definition of a function f(x) :

f '(x) = lim [h → 0] ( f(x + h) - f(x) ) / h

Then if x = a, we get

f '(a) = lim [h → 0] ( f(a + h) - f(a) ) / h

From here, it's easy to identify what each function and point should be:

(a) f (a + h) = (1 + h)¹ʹ³   →   f(x) = x ¹ʹ³ and a = 1

(that's a 1/3 in the exponent)

(b) f (a + h) = cos(π + h)   →   f(x) = cos(x) and a = π

(c) f (a + h) = 5 (4 + h)⁵   →   f(x) = 5x ⁵ and a = 4

(d) f (a + h) = exp(4h) = exp(4 (0 + h))   →   f(x) = exp(4x) and a = 0

(where exp(x) = eˣ )

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.