If f(x) = x^2 + 1 and g(x) = 3x + 1, find [f(4)]^2 81 257 289

2 answers:

7 0

So g(x) doesn't come in
ok

remember pemdas
inside first
evaluate f(4)
f(4)=4^2+1=16+1=17
now we have

[17]^2=289

answer s 289

7 0

Answer:

289 is the answer

Step-by-step explanation:

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ankoles [38]

Answer:

D. undefined

General Formulas and Concepts:

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Trig Derivative: $\displaystyle \frac{d}{dx}[sinu] = u'cosu$

Derivatives of Parametrics: $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Step-by-step explanation:

Step 1: Define

$\displaystyle \frac{dx}{dt} = 5$

$\displaystyle \frac{dy}{dt} = sin(t^2)$

Step 2: Differentiate

[x Derivative] Basic Power Rule: $\displaystyle \frac{d^2x}{dt^2} = 0$
[y Derivative] Trig Derivative [Chain Rule]: $\displaystyle \frac{d^2y}{dt^2} = cos(t^2) \cdot \frac{d}{dt}[t^2]$
[y Derivative] Basic Power Rule: $\displaystyle \frac{d^2y}{dt^2} = cos(t^2) \cdot 2t^{2 - 1}$
[y Derivative] Simplify: $\displaystyle \frac{d^2y}{dt^2} = 2tcos(t^2)$
[Derivative] Rewrite: $\displaystyle \frac{d^2y}{dx^2} = \frac{2tcos(t^2)}{0}$