Mathematics, 15.10.2019 15:10, zalliagessnerp5fuj9

Good answers get brainliest. +40 points wrong answers get reported
9.05a

1. find the derivative of f(x) = 8x2 + 11x at x = 7.

a) 123
b) -147
c) 156
d) 189

2. find the derivative of f(x) = 3x + 8 at x = 4.

a) 1
b ) 3
c) 4
d) 8

3. find the derivative of f(x) = 5 divided by x at x = -1.

a) 5
b) 1
c) -1
d) -5

4. find the derivative of f(x) = negative 7 divided by x at x = -3.

a) 3 divided by 7
b) 9 divided by 7
c) 7 divided by 9
d) 7 divided by 3

5. find the limit of the function by using direct substitution.
limit as x approaches zero of quantity x squared minus eight.

a) 0
b) does not exist
c) -8
d) 8

6. find the limit of the function by using direct substitution.
limit as x approaches two of quantity x squared plus three x minus one.

a) 0
b) does not exist
c) 9
d) -9

7. find the limit of the function algebraically.
limit as x approaches zero of quantity negative six plus x divided by x to the fourth power.

a) 6
b) 0
c) -6
d) does not exist

8. find the limit of the function algebraically.
limit as x approaches zero of quantity nine plus x divided by x to the third power.

a) does not exist
b) 9
c) -9
d) 0

9. use the given graph to determine the limit, if it exists.
a coordinate graph is shown with a horizontal line crossing the y axis at four that ends at the open point 2, 4, a closed point at 2, 1, and another horizontal line starting at the open point 2, -2.

find limit as x approaches two from the left of f of x. and limit as x approaches two from the right of f of x.

a) 4; -2
b) 1; 1
c) does not exist; does not exist
d) -2; 4

10. use the given graph to determine the limit, if it exists.
a coordinate graph is shown with a downward sloped line crossing the y axis at two that ends at the open point 2, 1.3, a closed point at 2, -1, and an upward sloped line starting at the open point 2, 4.

find limit as x approaches two from the left of f of x.

a) 1.3
b) -1
c) 4
d) 5

11. find the indicated limit, if it exists.
limit of f of x as x approaches 9 where f of x equals x plus 9 when x is less than 9 and f of x equals 9 minus x when x is greater than or equal to 9

a) the limit does not exist.
b) 18
c) 0
d) 9

12. find the indicated limit, if it exists.
limit of f of x as x approaches 1 where f of x equals 1 minus x when x is less than 1, 8 when x equals 1, and x plus 7 when x is greater than 1

a) 8
b) the limit does not exist.
c) 0
d) 7

13. use graphs and tables to find the limit and identify any vertical asymptotes of limit of 1 divided by the quantity x minus 9 as x approaches 9 from the left .

a) -∞; x = -9
b) 1 ; no vertical asymptotes
c) ∞; x = -9
d) -∞; x = 9

14. the position of an object at time t is given by s(t) = -2 - 6t. find the instantaneous velocity at t = 2 by finding the derivative.

15. use graphs and tables to find the limit and identify any vertical asymptotes of limit of 1 divided by the quantity x minus 7 squared as x approaches 7 .

Answers: 3

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Answers

answered: AndiLizzi

1. Ans:(A) 123

Given function: f(x) = 8x^2 + 11x

The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2 + 11x)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2) + \frac{d}{dx}(11x)$
=> $\frac{d}{dx} f(x) = 2*8(x^{2-1}) + 11$
=> $\frac{d}{dx} f(x) = 16x + 11$

Now at x = 7:
$\frac{d}{dx} f(7) = 16(7) + 11$

=> $\frac{d}{dx} f(7) = 123$

2. Ans:(B) 3

Given function: f(x) =3x + 8

The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(3x + 8)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(8)$
=> $\frac{d}{dx} f(x) = 3*1 + 0$
=> $\frac{d}{dx} f(x) = 3$

Now at x = 4:
$\frac{d}{dx} f(4) = 3$ (as constant)

=>Ans: $\frac{d}{dx} f(4) =$ 3

3. Ans:(D) -5

Given function: $f(x) = \frac{5}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{5}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(5x^{-1})$
=> $\frac{d}{dx} f(x) = 5*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = -5x^{-2}$

Now at x = -1:
$\frac{d}{dx} f(-1) = -5(-1)^{-2}$

=> $\frac{d}{dx} f(-1) = -5 *\frac{1}{(-1)^{2}}$
=> Ans: $\frac{d}{dx} f(-1) = -5$

4. Ans:(C) 7 divided by 9

Given function: $f(x) = \frac{-7}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{-7}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(-7x^{-1})$
=> $\frac{d}{dx} f(x) = -7*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = 7x^{-2}$

Now at x = -3:
$\frac{d}{dx} f(-3) = 7(-3)^{-2}$

=> $\frac{d}{dx} f(-3) = 7 *\frac{1}{(-3)^{2}}$
=> Ans: $\frac{d}{dx} f(-3) = \frac{7}{9}$

5. Ans:(C) -8

Given function:
f(x) = x^2 - 8

Now if we apply limit:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 - 8)$

=> $\lim_{x \to 0} f(x) = (0)^2 - 8$
=> Ans: $\lim_{x \to 0} f(x) = - 8$

6. Ans:(C) 9

Given function:
f(x) = x^2 + 3x - 1

Now if we apply limit:
$\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^2 + 3x - 1)$

=> $\lim_{x \to 2} f(x) = (2)^2 + 3(2) - 1$
=> Ans: $\lim_{x \to 2} f(x) = 4 + 6 - 1 = 9$

7. Ans:(D) doesn't exist.

Given function: $f(x) = -6 + \frac{x}{x^4}$
In this case, even if we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

Check:
$f(x) = -6 + \frac{x}{x^4} \\ f(x) = -6 + \frac{1}{x^3} \\ f(x) = \frac{-6x^3 + 1}{x^3} \\ Rationalize: \\ f(x) = \frac{-6x^3 + 1}{x^3} * \frac{x^{-3}}{x^{-3}} \\ f(x) = \frac{-6x^{3-3} + x^{-3}}{x^0} \\ f(x) = -6 + \frac{1}{x^3} \\ Same$

If you apply the limit, answer would be infinity.

8. Ans:(A) Doesn't Exist.

Given function: $f(x) = 9 + \frac{x}{x^3}$
Same as Question 7
If we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

9, 10.
Please attach the graphs. I shall amend the answer. :)

11. Ans:(A) Doesn't exist.

First We need to find out: $\lim_{x \to 9} f(x)$ where,
$f(x) = \left \{ {{x+9, ~~~~~x \textless 9} \atop {9- x,~~~~~x \geq 9}} \right.$

If both sides are equal on applying limit then limit does exist.

Let check:
If x $\textless$ 9: answer would be 9+9 = 18
If x $\geq$ 9: answer would be 9-9 = 0

Since both are not equal, as $18 \neq 0$ , hence limit doesn't exist.

12. Ans:(B) Limit doesn't exist.

Find out: $\lim_{x \to 1} f(x)$ where,

$f(x) = \left \{ {{1-x, ~~~~~x \textless 1} \atop {x+7,~~~~~x \textgreater 1} } \right. \\ and \\ f(x) = 8, ~~~~~ x=1$

If all of above three are equal upon applying limit, then limit exists.

When x < 1 -> 1-1 = 0
When x = 1 -> 8
When x > 1 -> 7 + 1 = 8

ALL of the THREE must be equal. As they are not equal. 0 $\neq$ 8; hence, limit doesn't exist.

13. Ans:(D) -∞; x = 9

f(x) = 1/(x-9).

Table:

x f(x)=1/(x-9)

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

8.9 -10

8.99 -100

8.999 -1000

8.9999 -10000

9.0 -∞

Below the graph is attached! As you can see in the graph that at x=9, the curve approaches but NEVER exactly touches the x=9 line. Also the curve is in downward direction when you approach from the left. Hence, -∞, x =9 (correct)

14. Ans: -6

s(t) = -2 - 6t

Inst. velocity = $\frac{ds(t)}{dt}$

Therefore,

$\frac{ds(t)}{dt} = \frac{ds(t)}{dt}(-2-6t) \\ \frac{ds(t)}{dt} = 0 - 6 = -6$

At t=2,

Inst. velocity = -6

15. Ans: +∞, x =7

f(x) = 1/(x-7)^2.

Table:

x f(x)= 1/(x-7)^2

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

6.9 +100

6.99 +10000

6.999 +1000000

6.9999 +100000000

7.0 +∞

Below the graph is attached! As you can see in the graph that at x=7, the curve approaches but NEVER exactly touches the x=7 line. The curve is in upward direction if approached from left or right. Hence, +∞, x =7 (correct)

-i

Good answers get brainliest. +40 points wrong answers get reported 9.05a 1. find the derivative of f

answered: Guest

step-by-step explanation:

answered: Guest

the largest number of the four options is 1.8 x 10-2

you have it in scientific notation.

1.80 x 10-6 = 0.00000180

1.80 x 10-4 = 0.000180 (bigger, or "less small", than the former one)

1.8 x 10-2 = 0.0180. as you see. the "less small", or the biggest one of them all.

as you have negative exponent, it means you are dividing 1.8 by 100 (i.e. moving the dot towards left, two times).

if you had positive exponent for the scientific notation, power of tenths, you would be multiplying times 100 (moving the dot towards right) as below:

1.8 x (10)2 = 180

so. if you have negative exponent in scientific notation, move the dot toward left. if it is positive, toward right as many times the exponent tells you.