Mathematics, 15.10.2019 15:10 zalliagessnerp5fuj9

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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

Answer from: 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