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2 years ago

More points for easy questions

Mathematics

2 answers:

Dmitrij [34]2 years ago

8 0

Answer:

sum ; cube

Step-by-step explanation:

(6x)³ is the cube of 6x

(6x)³ + 2 is the sum of cube of 6x and 2

Veseljchak [2.6K]2 years ago

6 0

Answer:

"the sum of the cube of 6x and 2"

Step-by-step explanation:

We see that only the expression of 6x is being cubed, not the entire expression of (6x)³ + 2.

The first blank must apply to both these terms 6x and 2, so the word can't be cubed. Instead, because we're finding the sum of these two, the word should be "sum".

The second blank should apply to only the 6x because we've already described the relationship of 6x and 2. So, since we are cubing 6x, the word here should be "cube".

The final sentence should read:

"the sum of the cube of 6x and 2"

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2 years ago

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3 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.

3 0

2 years ago