Question 40 is the one I could use some help on for all parts. 33–42. Derivatives and tangent linesa. For the following functions and values of a, find f'(a).b. Determine an equation of the line tangent to the graph of f at thepoint (a, f(a)) for the given value of a.33. f(x)= 8x; a-334. f(x) = x²; a = 335. f(x) = 4x² + 2x; a = -2 36. f(x) = 2x³; a = 10137. f(x) =11Via438. f(x)a = 139. f(x) = V2x + 1; a = 440. f(x) = V3x; a = 121; ax + 5141. f(x)542. f(x)2, a =3x – 1aх-

(40) The given function is: f(x)=3x at a = 12 (a) To find the derivative, apply the power rule…

Answered: 40. f(x) = VЗx; а — 12

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question 40 is the one I could use some help on for all parts.

33–42. Derivatives and tangent lines
a. For the following functions and values of a, find f'(a).
b. Determine an equation of the line tangent to the graph of f at the
point (a, f(a)) for the given value of a.
33. f(x)
= 8x; a
-3
34. f(x) = x²; a = 3
35. f(x) = 4x² + 2x; a = -2 36. f(x) = 2x³; a = 10
1
37. f(x) =
1
1
Via
4
38. f(x)
a = 1
39. f(x) = V2x + 1; a = 4
40. f(x) = V3x; a = 12
1
; a
x + 5
1
41. f(x)
5
42. f(x)
2
, a =
3x – 1a
х
-

Expert Solution

Step 1

(40)

The given function is:

$f (x) = \sqrt{3 x}$ at a = 12

(a)

To find the derivative, apply the power rule

$\frac{d}{d x} (x^{n}) = n x^{n - 1}$

So,

$\begin{array}{rcl} f' (x) & = & \frac{d}{d x} (\sqrt{3 x}) \\ = & \sqrt{3} \frac{d}{d x} \sqrt{x} \\ = & \sqrt{3} (\frac{1}{2} x^{\frac{1}{2} - 1}) \\ = & \frac{\sqrt{3}}{2 \sqrt{x}} \end{array}$

At x = a = 12,

$\begin{array}{rcl} f' (12) & = & \frac{\sqrt{3}}{2 \sqrt{12}} \\ = & \frac{1}{2} \sqrt{\frac{3}{12}} \\ = & \frac{1}{2} \sqrt{\frac{1}{4}} \\ = & \frac{1}{2 (2)} \\ = & \frac{1}{4} \end{array}$