Find all real zeros of the function y = 3x + 1.

Mathematics

1 answer:

murzikaleks [220]3 years ago

4 0

Answer:

$-\dfrac{1}{3}$

Step-by-step explanation:

The given function is

$y=3x+1$

We need to find all real zeros of the given function.

Equate y=0, to find all real zeros of the given function.

$3x+1=0$

Subtract 1 from both sides.

$3x+1-1=0-1$

$3x=-1$

Divide both sides by 3.

$\dfrac{3x}{3}=\dfrac{-1}{3}$

$x=-\dfrac{1}{3}$

Therefore, the real zero of the function is -1/3.

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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.