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jeka94
3 years ago
14

Write the cubic function.

Mathematics
1 answer:
creativ13 [48]3 years ago
4 0

Answer:

Step-by-step explanation:

hello :

let f this function

the roots of this function are : -1 , 0 ,6   so ;    f(x) = ax(x+1)(x-6)

now calculate : a  ....but this graph passes by : (1, -10)  means f(1) = -10

put : x=1   and  f(x) = -10

-10 = a(1)(1+1)(1-6)

-10a = -10    so : a= 1

conclusion :  

<h2> f(x) = x(x+1)(x-6)</h2>

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Answer:

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Step-by-step explanation:

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Answer:

see explanation

Step-by-step explanation:

There are 2 possible approaches to differentiating these.

Expand the factors and differentiate term by term, or

Use the product rule for differentiation.

I feel they are looking for use of product rule.

Given

y = f(x). g(x) , then

\frac{dy}{dx} = f(x).g'(x) + g(x).f'(x) ← product rule

(a)

y = (2x - 1)(x + 4)²

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g(x) = (x + 4)²

g'(x) = 2(x + 4) × \frac{d}{dx} (x + 4) ← chain rule

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Then

\frac{dy}{dx} = (2x - 1). 2(x + 4) + (x + 4)². 2

    = 2(2x - 1)(x + 4) + 2(x + 4)² ← factor out 2(x + 4) from each term

    = 2(x + 4) (2x - 1 + x + 4)

    = 2(x + 4)(3x + 3) ← factor out 3

    = 6(x + 4)(x + 1)

--------------------------------------------------------------------------

(b)

y =  x(x² - 1)³

f(x) = x ⇒ f'(x) = 1

g(x) = (x² - 1)³

g'(x) = 3(x² - 1)² × \frac{d}{dx} (x² - 1) ← chain rule

        = 3(x² - 1)² × 2x

        = 6x(x² - 1)²

Then

\frac{dy}{dx} = x. 6x(x² - 1)² + (x² - 1)³. 1

    = 6x²(x² - 1)² + (x² - 1)³ ← factor out (x² - 1)²

    = (x² - 1)² (6x² + x² - 1)

     = (x² - 1)²(7x² - 1)

----------------------------------------------------------------------

(c)

y = (x² - 1)(x³ + 1)

f(x) = x² - 1 ⇒ f'(x) = 2x

g(x) = (x³ + 1) ⇒ g'(x) = 3x²

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\frac{dy}{dx} = (x² - 1). 3x² + (x³ + 1), 2x

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   = x[3x(x² - 1) + 2(x³ + 1) ]

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g(x) = (x² + 4)²

g'(x) = 2(x² + 4) × \frac{d}{dx}(x² + 4) ← chain rule

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