The variables x and y are directly

1 answer:

7 0

Answer: When y = -4, x = -10.

Explanation:
The equation for direct variation is y = kx^n, where k is the constant of variation. n is the exponent above the x value, but since this problem doesn’t involve exponents greater than one, we can ignore it.

First, find k.
y = kx
6 = k*15
k = 6/15 or 2/5

Now, we can solve for any x or y input. Let’s find x when y = -4.
y = kx
-4 = 2/5 * x
(to cancel out the fraction, multiply each side of the equation by its reciprocal, 5/2)
x = -4 * 5/2
x = -20/2
x = -10

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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.