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3 years ago

9

What is the range of the function f(x) =4x-1 when the domain is -3 ,0,4 ?

1 answer:

dedylja [7]3 years ago

3 0

Answer:

The range of this function is {-13, -1, 15}.

Step-by-step explanation:

Evaluate f(x) = 4x - 1 at {-3, 0, 4}:

f(-3) = -13

f(0) = -1

f(4) = 15

The range of this function is {-13, -1, 15}.

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3 years ago

To find the equation of a line, we need the slope of the line and a point on the line. Since we are requested to find the equati

Ad libitum [116K]

Answer:

$m = \frac{1}{12}$

Step-by-step explanation:

Given

$(x,y) = (36,6)$

$f(x) = \sqrt x$ ----- the equation of the curve

Required

The slope of f(x)

The slope (m) is calculated using:

$m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

$(x,y) = (36,6)$ implies that:

$a = 36; f(a) = 6$

So, we have:

$m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

$m = \lim_{h \to 0} \frac{f(36 + h) - 6}{h}$

If $f(x) = \sqrt x$ ; then:

$f(36 + h) = \sqrt{36 + h}$

So, we have:

$m = \lim_{h \to 0} \frac{\sqrt{36 + h} - 6}{h}$

Multiply by: $\sqrt{36 + h} + 6$

$m = \lim_{h \to 0} \frac{(\sqrt{36 + h} - 6)(\sqrt{36 + h} + 6)}{h(\sqrt{36 + h} + 6)}$

Expand the numerator

$m = \lim_{h \to 0} \frac{36 + h - 36}{h(\sqrt{36 + h} + 6)}$

Collect like terms

$m = \lim_{h \to 0} \frac{36 - 36+ h }{h(\sqrt{36 + h} + 6)}$

$m = \lim_{h \to 0} \frac{h }{h(\sqrt{36 + h} + 6)}$

Cancel out h

$m = \lim_{h \to 0} \frac{1}{\sqrt{36 + h} + 6}$

$h \to 0$ implies that we substitute 0 for h;

So, we have:

$m = \frac{1}{\sqrt{36 + 0} + 6}$

$m = \frac{1}{\sqrt{36} + 6}$

$m = \frac{1}{6 + 6}$

$m = \frac{1}{12}$

<em>Hence, the slope is 1/12</em>

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3 years ago