b) Let function k be the function defined by k(x) = i ( f(x) ) . What is k' (n) ?%3D 2.Let function f be defined by f(x) = 4 cos (2x) – 2esin(3x).Let functions g and h be differentiable functions, where h(x) =g(x). The table belowgives function values of g and its derivative, g', at selected values of x .-5-3 -2 -1-2-4-5-8g(x)gʻ(x)10-3-4-5 -1-0.5Let function i be the function whose graph, consisting of five line segments, is shown in thefigure below.4 TyGraph of i-6-2452

Answered: Let function k be the function defined…

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

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ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter1: Functions

Section1.2: Functions Given By Tables

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b) Let function k be the function defined by k(x) = i ( f(x) ) . What is k' (n) ?
%3D

2.
Let function f be defined by f(x) = 4 cos (2x) – 2esin(3x).
Let functions g and h be differentiable functions, where h(x) =g(x). The table below
gives function values of g and its derivative, g', at selected values of x .
-5
-3 -2 -1
-2
-4
-5
-8
g(x)
gʻ(x)
10
-3
-4-5 -1
-0.5
Let function i be the function whose graph, consisting of five line segments, is shown in the
figure below.
4 Ty
Graph of i
-6
-2
452

Expert Solution

Step 1

$Given that, f (x) = 4 \cos (2 x) - 2 e^{\sin (3 x)} f (π) = 4 \cos (2 π) - 2 e^{\sin (3 π)} = 4 (1) - 2 e^{0} = 2$

$f' (x) = \frac{d f}{d x} = \frac{d}{d x} (4 \cos (2 x) - 2 e^{\sin (3 x)}) = - 8 \sin (2 x) - 6 e^{\sin (3 x)} \cos (3 x) f' (π) = - 8 \sin (2 π) - 6 e^{\sin (3 x)} \cos (3 π) = - 8 (0) - 6 e^{0} (- 1) = 0 + 6 = 6$

$Thus, f (π) = 2, f' (π) = 6$

Step 2

$For 0 \leq x \leq 3, i (x) is a straight line passing through the points (0, 0) and (3, - 1) . Thus, i (x) = - \frac{1}{3} x, 0 \leq x \leq 3$

$For 0 \leq x \leq 3, i' (x) = \frac{d}{d x} (- \frac{1}{3} x) = - \frac{1}{3} Thus, i' (2) = - \frac{1}{3}$

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