questions for my test review please help

Question

Comments

What need for 10 th question. please provide.

Answer 1

8)

a)

f(x) = (3x² - 4)⁴

Take derivative both sides with respect to x.

f '(x) = 4 (3x² - 4)³ d/dx (3x² - 4)

       = 4 (3x² - 4)³  (6x)

       = 24x(3x² - 4)³

Answer 2

8)b) The function

Apply quotient rule in derivatives

.

Answer 3

9) The function

Use the limit definition of the derivative

.

Comments

I don't understand the algebra for the last 3 step. How can you combine them. how it becomes 2?

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Also I think you are wrong, when the numberator was 2h, limit of h goes to 0, so 2 times 0 is 0, why is it still 2? So if you are wrong, what is the right answer?

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It's simple.

From the addition property a + a = 2a.

In the same way, the algebra explanation in last three steps in above solution.

Cancel common terms.

Back substitute

.

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Cancel the term h in numarator and denominator.

In the next step, apply the limits.

limit of h goes to 0.

Substitute 0 for h.

And fallow the above solution.

The Derivative of is is right answer.

Answer 4

(7)

Step 1 :

Given function is  : (2x³ - 0.6x² - 8.6x - 6) / (3x³ - 0.6x² - 12.24x - 8.64)

Numerator: 2x³ - 0.6x² - 8.6x - 6

Denominator: 3x³ - 0.6x² - 12.24x - 8.64

Identify Rational Zeros :

The Rational Zero Theorem can be used for finding the some possible zeros to test.

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation

a_nx ⁿ + a_n – 1x ^{n – 1} + ... + a₁x + a₀ = 0, then p is a factor of a₀ and q is a factor if a_n.

Numerator is 2x ³- 0.6x ²- 8.6x - 6 = 0.

If p/q is a rational zero, then p is a factor of 6 and q is a factor of 1.

The possible values of p  are   ± 1, ± 2, and ± 3.

The possible values for q are   ± 1, ± 2

By the Rational Roots Theorem, the only possible rational roots are, p / q = ± 1,   ± 2,  ± 3 , ±1/2 , ± 3/2

Make a table for the synthetic division and test possible real zeros.

p/q	1	-0.6	- 8.6	-6
1	1	-0.4	-9	-15
- 1	1	-1.6	- 6	0

Since, f(1) = 0, x = 1 is a zero. The depressed polynomial is  x ² - 1.6x - 6 = 0.

⇒ x = -1.2 and x = 2.5

Therefore, tthe roots of the function are x = -1, x = - 1.2, and x = 2.5

Denominator is 3x ³- 0.6x ²- 12.24x - 8.64

Substitute -1  in Denominator :

3(-1)³ - 0.6(-1)² - 12.24(-1) - 8.64

= - 3 - 0.6 + 12.24 - 8.64

= 0

Synthetic substitution p/q = -1

Make a table for the synthetic division and test possible real zeros.

p/q	3	-0.6	- 12.24	-8.64
-1	3	-3.6	-8.64	0

Since, f(-1) = 0, x = -1 is a zero. The depressed polynomial is  3x ² - 3.6x - 8.64 = 0.

⇒ x = -1.2 and x = 2.4

Therefore, the roots of the function are x = -1, x = - 1.2, and x = 2.4

Step 2 :

Given function is :

Step 3 :

To find a horizontal asymptote,

The degree of the numerator is the same as the degree of the denominator.

In this case, then the horizontal asymptote is a/d =1/1 = 1.

(a = the coefficient of the highest degree in the numerator

d = the coefficient of the highest degree in the denominator.)

To find a vertical asymptote,

set the denominator equal to 0 and solve for x

x - 2.4 = 0

x = 2.4

To find Removable discontinuity,

Canceled factors are ( x +1 )  and  (  x + 1.2 )

( x +1 )  = 0 and  (  x + 1.2 ) = 0

x  = -1  and x = -1.2

To find Non - Removable discontinuities,

set the denominator equal to 0 and solve for x

x - 2.4 = 0

x = 2.4

Solution :

Horizontal asymptote : 1

Vertical asymptote x = 2.4

Removable discontinuities  x  = -1  and x = -1.2

Non - Removable discontinuities x = 2.4

 

Comments

For number 7, how would I be able to do it on a calculator?

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We don't know what kind of calculator you are using .

We will help you , if you provide the type of calculator that your using .

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I am using the Ti-89 tiatinium