Model the pair of situations with exponential functions f and g. Find the approximate value of x that makes f(x)= g(x).

Mathematics

1 answer:

Law Incorporation [45]3 years ago

6 0

I like the cat on your pfp LOL sorry i couldn’t help tho

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Help me plz plllzzzz

gavmur [86]

Whenever you multiply by a negative number.

Hopefully that helps you a little.

7 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

3 years ago

Use the formula d=rt Find t for r = 48.3 m/h and d = 545.79 m.

djyliett [7]

D = 545.79
r = 48.3

Plug in numbers to corresponding variables

545.79 = t(48.3)

Isolate the t, divide 48.3 from both sides

(545.79)/48.3 = 48.3t/48.3

t = 545.79/48.3
t = 11.3

t = A) 11.3 h

hope this helps

8 0

3 years ago

Which function is invertible? select EACH correct answer.

Elden [556K]

Step-by-step explanation:

this two...? i think have a nice day!