please help me please i am on a test

Question

Question 1
Find the derivative of f(x) = -10x2 + 4x at x = 11.
-216
-196
-176
-363

Question 2
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
Find the derivative of f(x) = 7x + 9 at x = 6.
7
9
6
0

Question 4
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
Find the derivative of f(x) = 3 divided by x at x = 1.
-3
-1
1
3

Question 6
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
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
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
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
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
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
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
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
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

 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

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

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