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3 years ago

Which second degree polynomial function f(x) has a lead coefficient of 3 and roots 4 and 1?

Mathematics

2 answers:

Stella [2.4K]3 years ago

7 0

Answer:

f(x) = 3x² - 15x + 12

Step-by-step explanation:

Given f(x) has roots x = a and x = b, then

(x - a) and (x - b) are the factors

f(x) is then the product of the factors

f(x) = a(x - a)(x - b) ← where a is a multiplier

Given roots are x = 4 and x = 1, then

(x - 4) and (x - 1) are the factors

With a = 3, then

f(x) = 3(x - 4)(x - 1) ← expand factors using FOIL

= 3(x² - 5x + 4) ← distribute

= 3x² - 15x + 12

Sophie [7]3 years ago

5 0

Answer:

f(x) = 3x^2 - 15x + 12

Step-by-step explanation:

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Morgarella [4.7K]

Answer:

$f'(x)=-\frac{2}{x^\frac{3}{2}}$

Step-by-step explanation:

So we have the function:

$f(x)=\frac{4}{\sqrt x}$

And we want to find the derivative using the limit process.

The definition of a derivative as a limit is:

$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Therefore, our derivative would be:

$\lim_{h \to 0}\frac{\frac{4}{\sqrt{x+h}}-\frac{4}{\sqrt x}}{h}$

First of all, let's factor out a 4 from the numerator and place it in front of our limit:

$=\lim_{h \to 0}\frac{4(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x})}{h}$

Place the 4 in front:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}$

Now, let's multiply everything by (√(x+h)(√(x))) to get rid of the fractions in the denominator. Therefore:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x})$

Distribute:

$=4\lim_{h \to 0}\frac{({\sqrt{x+h}\sqrt x})\frac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\frac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}$

Simplify: For the first term on the left, the √(x+h) cancels. For the term on the right, the (√(x)) cancel. Thus:

$=4 \lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }$

Now, multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x+h)). Thus:

$= 4\lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }(\frac{\sqrt x +\sqrt{x+h})}{\sqrt x +\sqrt{x+h})}$

The numerator will use the difference of two squares. Thus:

$=4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Simplify the numerator:

$=4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Both the numerator and denominator have a h. Cancel them:

$=4 \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Now, substitute 0 for h. So:

$=4 ( \frac{-1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})})$

Simplify:

$=4( \frac{-1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})})$

(√x)(√x) is just x. (√x)+(√x) is just 2(√x). Therefore:

$=4( \frac{-1}{(x)(2\sqrt{x})})$

Multiply across:

$= \frac{-4}{(2x\sqrt{x})}$

Reduce. Change √x to x^(1/2). So:

$=-\frac{2}{x(x^{\frac{1}{2}})}$

Add the exponents:

$=-\frac{2}{x^\frac{3}{2}}$

And we're done!

$f(x)=\frac{4}{\sqrt x}\\f'(x)=-\frac{2}{x^\frac{3}{2}}$

5 0

2 years ago

Which best describes the number 0.25?

Doss [256]

Answer:

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Create a recursive formula for this sequence (linear)

svetoff [14.1K]

the answer is 3n-2 please do you get it

4 0

2 years ago

2) Line segment MK has endpoints at (2, 3) and (5, ?4). Segment M'K' is the reflection of MK over the y-axis. Which statement de

Marianna [84]

Answer:

Option C -M'K' is the same length as MK

Step-by-step explanation:

Given : Line segment MK has endpoints at (2, 3) and (5,4)

M'K' is the reflection of MK over the y-axis

By definition of reflection: reflection of point (x,y) across the the y-axis is the point (-x,y)

which implies M'K' has end points (-2,3) and (-5,4)

Now, we find the length of MK

let $(x_1,y_1)=(2,3)\\\\(x_2,y_2)=(5,4)$

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

⇒ $d=\sqrt{(2-5)^2+(4-3)^2}$

⇒ $d=\sqrt{9+1}$

⇒ $d=\sqrt{10}$ ....(1)

Now, we find the length of M'K'

let $(x_1^{'},y_1^{'})=(-2,3)\\\\(x_2^{'},y_2^{'})=(-5,4)$

$d^{'}=\sqrt{(x_2^{'}-x_1^{'})^2+(y_2^{'}-y_1^{'})^2}$

⇒ $d^{'}=\sqrt{(-2+5)^2+(3-4)^2}$

⇒ $d^{'}=\sqrt{9+1}$

⇒ $d^{'}=\sqrt{10}$ .....(2)

from (1) and (2) we simply show that the length of MK and M'K' is equal

we can also refer the figure attached for reflection of MK and M'K'

therefore, Option C is correct

3 0

3 years ago

Read 2 more answers

Hector has 7/8 of a pound of leftover Halloween candy. He plans on giving it away to his friends. If he plans on giving each per

ale4655 [162]

Answer: 14

Step-by-step explanation:

1/16x = 7/8

x = 7/8*16/1 = 14

3 0

2 years ago

Read 2 more answers