Find f(-2) for f(x) = 2(4)^x O A.1/16 O B.1/8 O C. -16 OD. - 32

Mathematics

1 answer:

scoray [572]2 years ago

5 0

Answer:

B. 1/8

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Function Notation

Step-by-step explanation:

Step 1: Define

$\displaystyle f(x) = 2(4)^x\\f(-2) \ is \ x = -2$

Step 2: Evaluate

Substitute in x: $\displaystyle f(-2) = 2(4)^{-2}$
Exponents: $\displaystyle f(-2) = 2(\frac{1}{16})$
Multiply: $\displaystyle f(-2) = \frac{1}{8}$

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antiseptic1488 [7]

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4 0

2 years ago

F(x) = lnx/x

My name is Ann [436]

$f(x) = \frac{lnx}{x}$

(a) $f'(x) = \frac{d}{dx}[\frac{lnx}{x}]$
Using the quotient rule:
$f'(x) = \frac{\frac{1}{x} \cdot x - lnx}{x^{2}}$
$f'(x) = \frac{1 - lnx}{x^{2}}$

For maximum, f'(x) = 0;
$\frac{1 - lnx}{x} = 0$
$lnx = 1, x = e$

(b) Deduce:
 $e^{x} \geq x^{e}, x > 0$

Soln: Since x = e is the greatest value, then f(e) ≥ f(x) > f(0)
$\frac{ln(e)}{e} \geq \frac{lnx}{x}$
$\frac{1}{e} \geq \frac{lnx}{x}$ , since ln(e) is simply equal to 1

Now, since x > 0, then we don't have to worry about flipping the signs when multiplying by x.
$\frac{x}{e} \geq lnx$
$x \geq elnx$
$x \geq ln(x^{e})$

Taking the exponential to both sides will cancel with the natural logarithmic function in the right hand side to produce:
$e^{x} \geq e^{ln(x^{e}})$
$e^{x} \geq x^{e}$ , as required.