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2 years ago

Question in image, it asks to show your work

Mathematics

1 answer:

maxonik [38]2 years ago

8 0

Answer:

• f(g(x)) = $\sqrt{x^{2} + 1}$ + 8

• g(f(x)) = 16 $\sqrt{x}$ + x + 65

Step-by-step explanation:

• f(g(x)) = f(x² + 1)

= $\sqrt{x^{2} + 1}$ + 8

• g(f(x)) = g( $\sqrt{x + 8}$ )

= $(\sqrt{x } + 8)^{2}$ + 1

= x + 16 $\sqrt{x}$ + 64 + 1

= 16 $\sqrt{x}$ + x + 65

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Help with b please. thank you

erastovalidia [21]

Answer:

See explanation.

General Formulas and Concepts:

Algebra I

Terms/Coefficients
Factoring

Algebra II

Polynomial Long Division

Pre-Calculus

Parametrics

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Parametric Differentiation: $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle x = 2t - \frac{1}{t}$

$\displaystyle y = t + \frac{4}{t}$

Step 2: Find Derivative

[x] Differentiate [Basic Power Rule and Quotient Rule]: $\displaystyle \frac{dx}{dt} = 2 + \frac{1}{t^2}$
[y] Differentiate [Basic Power Rule and Quotient Rule]: $\displaystyle \frac{dy}{dt} = 1 - \frac{4}{t^2}$
Substitute in variables [Parametric Derivative]: $\displaystyle \frac{dy}{dx} = \frac{1 - \frac{4}{t^2}}{2 + \frac{1}{t^2}}$
[Parametric Derivative] Simplify: $\displaystyle \frac{dy}{dx} = \frac{t^2 - 4}{2t^2 + 1}$
[Parametric Derivative] Polynomial Long Division: $\displaystyle \frac{dy}{dx} = \frac{1}{2} - \frac{7}{2(2t^2 - 1)}$
[Parametric Derivative] Factor: $\displaystyle \frac{dy}{dx} = \frac{1}{2} \bigg( 1 - \frac{9}{2t^2 + 1} \bigg)$

Here we see that if we increase our values for t, our derivative would get closer and closer to 0.5 but never actually reaching it. Another way to approach it is to take the limit of the derivative as t approaches to infinity. Hence $\displaystyle \frac{dy}{dx} < \frac{1}{2}$ .