Question

Question 1(Multiple Choice Worth 6 points)

Find the derivative of f(x) = -10x2 + 4x at x = 11.
-216
-196
-176
-363

Question 2(Multiple Choice Worth 6 points)
Find the limit of the function by using direct substitution.
limit as x approaches four of quantity x squared plus three x minus one
Does not exist
-27
0
27

Question 3(Multiple Choice Worth 6 points)
Find the derivative of f(x) = 7x + 9 at x = 6.
7
9
6
0

Question 4(Multiple Choice Worth 7 points)
Find the indicated limit, if it exists.
limit of f of x as x approaches 0 where f of x equals 7 minus x squared when x is less than 0, 7 when x equals 0, and 10 x plus 7 when x is greater than 0
3
10
7
The limit does not exist.

Question 5(Multiple Choice Worth 6 points)
Find the derivative of f(x) = 3 divided by x at x = 1.
-3
-1
1
3

Question 6(Multiple Choice Worth 7 points)
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, negative 3.
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..
-3; 4
1; 1
4; -3
Does not exist; does not exist

Question 7(Multiple Choice Worth 7 points)
Find the indicated limit, if it exists.
limit of f of x as x approaches negative 1 where f of x equals x plus 1 when x is less than negative 1 and f of x equals 1 minus x when x is greater than or equal to negative 1
-1
2
The limit does not exist.
0

Question 8(Multiple Choice Worth 7 points)
Use graphs and tables to find the limit and identify any vertical asymptotes of limit of 1 divided by the quantity x minus 7 as x approaches 7 from the left.
-∞ ; x = 7
∞ ; x = -7
-∞ ; x = -7
1 ; no vertical asymptotes

Question 9(Multiple Choice Worth 7 points)
Use the given graph to determine the limit, if it exists.
A coordinate graph is shown with an upward sloped line crossing the y axis at the origin that ends at the open point 3, 1.5, a closed point at 3, 7, and a horizontal line starting at the open point 3, 2.
Find limit as x approaches three from the left of f of x..
7
1.5
Does not exist
2

Question 10(Multiple Choice Worth 6 points)
Find the limit of the function algebraically.
limit as x approaches zero of quantity x cubed plus one divided by x to the fifth power.
0
-9
Does not exist
9

Question 11(Multiple Choice Worth 6 points)
Find the derivative of f(x) = negative 3 divided by x at x = -4.
3 divided by 4
3 divided by 16
16 divided by 3
4 divided by 3

Question 12(Multiple Choice Worth 6 points)
Find the limit of the function by using direct substitution.
limit as x approaches zero of quantity x squared minus two.
2
-2
Does not exist
0

Question 13(Multiple Choice Worth 6 points)
Find the limit of the function algebraically.
limit as x approaches four of quantity x squared minus sixteen divided by quantity x minus four.
Does not exist
4
1
8

Question 14 (Essay Worth 9 points)
The position of an object at time t is given by s(t) = -1 - 13t. Find the instantaneous velocity at t = 8 by finding the derivative.

Question 15 (Essay Worth 8 points) Use graphs and tables to find the limit and identify any vertical asymptotes of the function. limit of 1 divided by the quantity x minus 2 squared as x approaches 2

Answer 1

Answer :

Q1. f'(x)=-20x+4. When x=11, this is f'(11)=-20×11+4=-220+4=-216.

Q2. Lim(x➝4) of x²+3x-1. The expression can be evaluated by direct substitution without constraints: 16+12-1=27.

Q3. The derivative is 7 for all values of x because the function is linear with a constant slope.

Q4. The function is piecewise. We need the conditions that apply when x is close to zero. We need to look at x approaching from the negative side and the positive side. If the limits are not the same the limit does not exist. So, when x<0 we use 7-x². When x is just less than zero, this left-hand limit is 7. When x>0 we use 10x+7. When x is just bigger than zero, this right-hand limit is 7. We are not interested in x=0, we are only interested in the left- and right-hand limits, which happen to be the same. So the limit is 7.

Q5. f'(x)=-3/x². f'(1)=-3.

Q6. In all limit questions, you can always ignore what happens when x=limit, because it’s the approaches that matter. So x➝2⁻ means look at the graph on the left of the limit. The horizontal line at y=4 means the left limit is 4. x➝2⁺ means look at the graph on the right of the limit. This is where y=-3, so the right limit is -3. The answer is 4 and -3 for left and right limits respectively.

Q7. First find left and right limits: x➝-1⁻: use x<-1: f(x)=x+1, so left limit is 0; x➝-1⁺: use x>-1: f(x)=1-x: so right limit is 1-(-1)=2. The left and right limits are different, so the limit does not exist.

Q8. 1/(x-7). When x=7, this quantity is not defined and this is a vertical asymptote. When x➝7⁻ we can use a value of x slightly less than 7 and see what happens. First try x=6.99, then 1/(x-7)=-100. Now try x=6.999, then we get -1000. So it’s getting larger in magnitude, and it’s negative. An infinitely large negative value is -∞. So the limit as x approaches 7 from the left is -∞.

Q9. Ignore the closed point at (3,7). For x➝3⁻, we are on the sloping line, so the limit is 1.5.

Q10. As x approaches zero the numerator approaches 1. The denominator gets smaller so its reciprocal gets larger, and it can be positive or negative depending on how we approach zero. So we can’t define the limit, so it doesn’t exist.

Q11. Derivative is -(-3/x²)=3/x². When x=-4, this is 3/16, because (-4)²=16.

Q12. x²-2➝-2 when x➝0.

Q13. (x²-16)/(x-4)=(x-4)(x+4)/(x-4)➝x+4 because the common factor x-4 cancels. Note that the quantity does not exist at precisely x=4 but in the limit it approaches x+4=8 when x=4. So the answer is 8 for the limit.

Q14. The derivative is constant -13 and is independent of t, so the instantaneous velocity is -13.

Q15. 1/(x-2)² is always positive, so it doesn’t matter whether we look at the left or right limits. The result as x➝2 is ∞, which is an undefinable quantity. x=2 is a vertical asymptote.

Answer 2

ANSWER TO QUESTION 1

The given function is
`f(x)=-10x^2+4x`. We want to find the derivative of this function at
`x=11`.

The derivative of this function is given by,

`f'(x)=-20x+4`

We now have to substitute
`x=11` in to
`f'(x)=-20x+4` to obtain,

`f'(11)=-20(11)+4`

This implies that,

`f'(11)=-220+4`

`f'(11)=-216`

Ans: A

ANSWER TO Q2.

The given limit is
`\lim_(x \to 4)^( x^2+3x-1)`.

The function whose limit we are finding is a polynomial function. Since polynomial function are continuous everywhere, the limiting value is always equal to the functional value.

Thus,
`\lim_(x \to 4)^( x^2+3x-1)=f(4)`

This implies that,

Thus,
`\lim_(x \to 4)^( x^2+3x-1)=(4)^2+3(4)-1`.

`\lim_(x \to 4)^( x^2+3x-1)=16+12-1`

`\lim_(x \to 4)^( x^2+3x-1)=27`

Ans: A.

ANSWER TO Q3

The given function is
`f(x)=7x+9`.

We want to find the derivative of this function at
`x=6`.

We must first of all differentiate this function to obtain,

`f'(x)=7`

We can see that the derived function is constant, therefore value of
`x` will still give us
`7.`

This implies that,

`f'(6)=7`

Ans: A

ANSWER TO Q4.

See attachment

ANSWER TO Q5.

The given function is
`f(x)=(3)/(x)`.

To find the derivative of this function at,

`x=1`, we must first differentiate this function.

But let us rewrite the rational function as a power function so that it will be easier to differentiate using the power rule of differentiation.

That is,

`f(x)=3x^(-1)`.

We differentiate now to obtain,

`f'(x)=-1(3x^(-1-1))`

This implies that,

`f'(x)=-3x^(-2)`

`f'(x)=(-3)/(x^2)`

At
`x=1`

`f'(1)=(-3)/((1)^2)`

`f'(1)=-3`

Ans: A

ANSWER TO Q6

The graph that has been described in the question is shown in the diagram above,

We can see from the graph that as we approach
`x=2`, from the left, the y-values are approaching
`4`.

That is,

`\lim_(x \to 2^(-)) f(x)=4`

Also as we approach
`x=2` from the right, the y-values are approaching
`-3`.

That is,

`\lim_(x \to 2^(+)) f(x)=-3`.

Ans: D.

ANSWER TO Q7

The given piece-wise function is

`\lim_(x \to -1)  \left \{ {{x+1,if\:x\:<\:-1} \atop {1-x,if\:x\:\ge -1}} \right.`

Since
`x=-1 \in x\geq -1`

We evaluate the limit at
`x=-1` of
`f(x)=1-x`.

Since this is a polynomial function,
`\lim_(x \to -1)f(x)= 1-x=f(-1)`

This implies that,

`\lim_(x \to -1) f(x)=1-x=1--1`

`\lim_(x \to -1) f(x)=1-x=1--1`

`\lim_(x \to -1) f(x)=1-x=1+1`

`\lim_(x \to -1) f(x)=1-x=2`

Ans: B.

ANSWER TO Q8

The given function is
`f(x)=(1)/(x-7)`.

This is a rational function that is defined for all real values except,
`x-7=0`

Therefore the vertical asymptote is
`x=7`

We can see from the graph above that, as x approaches seven from the left, the function approaches negative infinity

That is
`\lim_(x \to 7^(-))  (1)/(x-7) =- \infty`.

Ans: A.

ANSWER TO Q9

The graph of the given piece-wise function is show as follows;

We can see from the graph that, as the x-values are approaching 3 from the left y-values are approaching
`1.5`.

Note the limit is not the same as the functional value in this case. Also limit in this case is the y-value we are approaching as we get closer and closer to 3, not necessarily at 3.

See graph

Ans: B

ANSWER TO Q10

The given function is
`f(x)=(x^3+1)/(x^5)`

We want to find the limiting value of this function as the x-values approaches zero.

Thus,

`\lim_(x \to 0) f(x)=(x^3+1)/(x^5)`

Since the function is not defined at
`x=0`.

We evaluate the one-sided limits as follows,

`\lim_(x \to 0^-) f(x)=(x^3+1)/(x^5)=- \infty`

`\lim_(x \to 0^+) f(x)=(x^3+1)/(x^5)=+ \infty`

Since the right hand limit is not equal to the left hand limit,

`\lim_(x \to 0) f(x)=(x^3+1)/(x^5)` Does Not Exist.

ANS: C

ANSWER TO Q11

The given function is

`f(x)=-(3)/(x)`

To find the derivative of this function at,

`x=-4`, we must first differentiate this function.

But let us rewrite the rational function as a power function so that it will be easier to differentiate using the power rule of differentiation.

That is,

`f(x)=-3x^(-1)`.

We differentiate now to obtain,

`f'(x)=-1(-3x^(-1-1))`

This implies that,

`f'(x)=3x^(-2)`

`f'(x)=(3)/(x^2)`

At
`x=-4`

`f'(1)=(3)/((-4)^2)`

`f'(1)=(3)/((16))`

Ans: B

ANSWER TO Q12

We want to find the limit of the function
`f(x)=x^2-2` as x approaches zero.

`\lim_(x \to 0) x^2-2`

This is a polynomial function, therefore

`\lim_(x \to 0) x^2-2=f(0)=0^2-2=-2`

Ans: B

SEE ATTACHMENT FOR ANSWER TO QUESTIONS

13, 14 and 15