Question

Show all of your work, even though the question may not explicitly remind you to do so. Clearly label any functions, graphs, tables, or other objects that you use. Justifications require that you give mathematical reasons, and that you verify the needed conditions under which relevant theorems, properties, definitions, or tests are applied. Your work will be scored on the correctness and completeness of your methods as well as your answers. Answers without supporting work will usually not receive credit. Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If your answer is given as a decimal approximation, it should be correct to three places after the decimal point. Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers for which f() is a real number Let f be an increasing function with f(0) = 2. The derivative of f is given by f'(x) = sin(TX) + 2? +3. (a) Find t" (-2) (b) Write an equation for the line tangent to the graph of y = ke at = 0. (c) Let I be the function defined by g(x) = f (V3x2 + 4). Find (2). (d) Let h be the inverse function of f. Find h' (2). Please respond on separate paper, following directions from your teacher.

Answer 1

Answer:

a. 5 b. $y = -\frac{3}{4}x + \frac{1}{2}$ c. 148.5 d. 1/7

Step-by-step explanation:

Here is the complete question

Show all of your work, even though the question may not explicitly remind you to do so. Clearly label any functions, graphs, tables, or other objects that you use. Justifications require that you give mathematical reasons, and that you verify the needed conditions under which relevant theorems, properties, definitions, or tests are applied. Your work will be scored on the correctness and completeness of your methods as well as your answers. Answers without supporting work will usually not receive credit. Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If your answer is given as a decimal approximation, it should be correct to three places after the decimal point. Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers for which f() is a real number Let f be an increasing function with f(0) = 2. The derivative of f is given by f'(x) = sin(πx) + x² +3. (a) Find f" (-2) (b) Write an equation for the line tangent to the graph of y = 1/f(x) at x = 0. (c) Let I be the function defined by g(x) = f (√(3x² + 4). Find g(2). (d) Let h be the inverse function of f. Find h' (2). Please respond on separate paper, following directions from your teacher.

Solution

a. f"(2)

f"(x) = df'(x)/dx = d(sin(πx) + x² +3)/dx = cos(πx) + 2x

f"(2) = cos(π × 2) + 2 × 2

f"(2) = cos(2π) + 4

f"(2) = 1 + 4

f"(2) = 5

b. Equation for the line tangent to the graph of y = 1/f(x) at x = 0

We first find f(x) by integrating f'(x)

f(x) = ∫f'(x)dx = ∫(sin(πx) + x² +3)dx = -cos(πx)/π + x³/3 +3x + C

f(0) = 2 so,

2 = -cos(π × 0)/π + 0³/3 +3 × 0 + C

2 = -cos(0)/π + 0 + 0 + C

2 = -1/π + C

C = 2 + 1/π

f(x) = -cos(πx)/π + x³/3 +3x + 2 + 1/π

f(x) = [1-cos(πx)]/π + x³/3 +3x + 2

y = 1/f(x) = 1/([1-cos(πx)]/π + x³/3 +3x + 2)

The tangent to y is thus dy/dx

dy/dx = d1/([1-cos(πx)]/π + x³/3 +3x + 2)/dx

dy/dx = -([1-cos(πx)]/π + x³/3 +3x + 2)⁻²(sin(πx) + x² +3)

at x = 0,

dy/dx = -([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)⁻²(sin(π × 0) + 0² +3)

dy/dx = -([1-cos(0)]/π + 0 + 0 + 2)⁻²(sin(0) + 0 +3)

dy/dx = -([1 - 1]/π + 0 + 0 + 2)⁻²(0 + 0 +3)

dy/dx = -(0/π + 2)⁻²(3)

dy/dx = -(0 + 2)⁻²(3)

dy/dx = -(2)⁻²(3)

dy/dx = -3/4

At x = 0,

y = 1/([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)

y = 1/([1-cos(0)]/π + 0 + 0 + 2)

y = 1/([1 - 1]/π + 2)

y = 1/(0/π + 2)

y = 1/(0 + 2)

y = 1/2

So, the equation of the tangent at (0, 1/2) is

$\frac{y - \frac{1}{2} }{x - 0} = -\frac{3}{4} \\y - \frac{1}{2} = -\frac{3}{4}x\\y = -\frac{3}{4}x + \frac{1}{2}$

c. If g(x) = f (√(3x² + 4). Find g'(2)

g(x) = f (√(3x² + 4) = [1-cos(π√(3x² + 4)]/π + √(3x² + 4)³/3 +3√(3x² + 4) + 2

g'(x) = [3xsinπ√(3x² + 4) + 18x(3x² + 4) + 9x]/√(3x² + 4)

g'(2) = [3(2)sinπ√(3(2)² + 4) + 18(2)(3(2)² + 4) + 9(2)]/√(3(2)² + 4)

g'(2) = [6sinπ√(12 + 4) + 36(12 + 4) + 18]/√12 + 4)

g'(2) = [6sinπ√(16) + 36(16) + 18]/√16)

g'(2) = [6sin4π + 576 + 18]/4)

g'(2) = [6 × 0 + 576 + 18]/4)

g'(2) = [0 + 576 + 18]/4)

g'(2) = 594/4

g'(2) = 148.5

d. If h be the inverse function of f. Find h' (2)

If h(x) = f⁻¹(x)

then h'(x) = 1/f'(x)

h'(x) = 1/(sin(πx) + x² +3)

h'(2) = 1/(sin(π2) + 2² +3)

h'(2) = 1/(sin(2π) + 4 +3)

h'(2) = 1/(0 + 4 +3)

h'(2) = 1/7

Answer 2

The function f'(x) is a trigonometric function

• The value of f''(2) is 5
• The equation of the tangent is y = -3/4x + 1/2
• The value of g'(2) is 148.500
• The value of h'(2) is 1/7

Part A: f"(2)

We have:

$f'(x) = \sin(\pi x) + x\² +3$

Differentiate with respect to xto get f''(x)

$f''(x)= \cos(\pi x) + 2x$

Substitute 2 for x

$f''(2) = \cos(2\pi) + 2(2)$

Evaluate cos(2π) and 2(2)

f''(2) = 1 + 4

Evaluate the sum

f''(2) = 5

Hence, the value of f''(2) is 5

Part B. Equation of tangent line to the graph of y = 1/f(x) at x = 0

Given that:

f(0) = 2

Recall that:

$f'(x) = \sin(\pi x) + x\² +3$

Integrate f'(x) with respect to x to calculate f(x)

$f(x) = \int f'(x)dx$

$f(x) = \int (sin(\pi x) + x\² +3)dx$

$f(x) = -\frac{\cos(\pi x)}{\pi} + \frac{x\³}{3} +3x + c$

Recall that f(0) = 2

So, we substitute 0 for x and 2 for f(x) to calculate c

$2 = -\frac{\cos(0\pi)}{\pi} + \frac{0\³}{3} +3 * 0 + c$

$2 = -\frac{1}{\pi} + c$

Make c the subject

$c =2 + \frac{1}{\pi}$

Substitute c = 2 + 1/π in f(x)

$f(x) = -\frac{\cos(\pi x)}{\pi} + \frac{x\³}{3} +3x + 2 + \frac{1}{\pi}$

Collect like terms

$f(x) = \frac{1}{\pi} - \frac{\cos(\pi x)}{\pi} + \frac{x\³}{3} +3x + 2$

Evaluate

$f(x) = \frac{1-\cos(\pi x)}{\pi} + \frac{x\³}{3} +3x + 2$

Recall that: y = 1/f(x)

So, we have:

$y = \frac 1{\frac{1-\cos(\pi x)}{\pi} + \frac{x\³}{3} +3x + 2}$

Calculate y when x = 0

$y = \frac 1{\frac{1-\cos(0)}{\pi} + \frac{0\³}{3} +3(0) + 2}$

$y = \frac{1}{\frac{1 - 1}{\pi} + 2}$

$y = \frac{1}{\frac{0}{\pi} + 2}$

y = 1/2

Differentiate with respect to x, to calculate dy/dx

$\frac{dy}{dx} = -(\frac{1-\cos(\pi x)}{\pi} + \frac{x\³}{3} +3x + 2)^{-2}} (\sin(\pi x) + x\² +3)$

Calculate the slope i.e. dy/dx when x = 0

$\frac{dy}{dx} = -(\frac{1-\cos(0)}{\pi} + \frac{0\³}{3} +3(0) + 2)^{-2}} (\sin(0) + 0\² +3)$

$\frac{dy}{dx} = -(\frac{0}{\pi} + 2)^{-2}} (3)$

$\frac{dy}{dx} = -(2)^{-2}} (3)$

Evaluate the exponent

$\frac{dy}{dx} = -\frac 14 * 3$

$\frac{dy}{dx} = -\frac 34$

Rewrite as:

$m= -\frac 34$

A linear equation is represented as:

y - y₁ = m(x - x₁)

So, we have:

y - 1/2 = -3/4(x - 0)

y - 1/2 = -3/4x

Add 1/2 to both sides

y = -3/4x + 1/2

Hence, the equation of the tangent is y = -3/4x + 1/2

Part C: Find g'(2)

In (b), we have:

$f(x) = \frac{1-\cos(\pi x)}{\pi} + \frac{x\³}{3} +3x + 2$

Substitute √(3x² + 4) for x in f(x)

$f(\sqrt{3x^2 + 4}) = \frac{1-\cos(\pi \sqrt{3x^2 + 4})}{\pi} + \frac{\sqrt{3x^2 + 4}\³}{3} +3\sqrt{3x^2 + 4} + 2$

So, we have:

$g(x) = \frac{1-\cos(\pi \sqrt{3x^2 + 4})}{\pi} + \frac{\sqrt{3x^2 + 4}\³}{3} +3\sqrt{3x^2 + 4} + 2$

Differentiate, to calculate g'(x)

$g'(x) = \frac{3xsin\pi\sqrt{3x\² + 4} + 18x(3x\² + 4) + 9x}{\sqrt{3x\² + 4}}$

Substitute 2 for x

$g'(x) = \frac{3*2 \sin\pi\sqrt{3(2)\² + 4} + 18x(3(2)\² + 4) + 9(2)}{\sqrt{3(2)\² + 4}}$

Evaluate

g'(2) = 148.500

Hence, the value of g'(2) is 148.500

Part D: Find h'(2), if h is the inverse function of f.

$f'(x) = \sin(\pi x) + x\² +3$

This means that:

$h'(x) = \frac{1}{f'(x)}$

The equation becomes

$h'(x) = \frac{1}{\sin(\pi x) + x\² +3}$

Substitute 2 for x

$h'(2) = \frac{1}{\sin(2\pi) + 2\² +3}$

Evaluate

$h'(2) = \frac 17$

Hence, the value of h'(2) is 1/7

Read more about equation of tangent line at:

https://brainly.com/question/10597313