Respuesta :
(1) Ans: Option (B) -69
Given function: [tex]f(x) = -4x^2 + 11x[/tex]
The derivative of f(x) with respect to x is:
[tex] \frac{df}{dx} = -4 * 2x + 11 [/tex] --- (1)
Plug the value of x = 10 in (1)
(1) => [tex]\frac{df}{dx} |_{x=10} = -4 * 2(10) + 11 = -69[/tex]
Hence the correct answer is Option (B) -69
(2) Ans: Option (C) 8
Given function: [tex]f(x) = 8x + 4[/tex]
The derivative of f(x) with respect to x is:
[tex] \frac{df}{dx} = 8 [/tex] --- (1)
Plug the value of x = 9 in (1)
(1) => [tex]\frac{df}{dx} |_{x=9} = 8[/tex]
Hence the correct answer is Option (C) 8
(3) Ans: Option (B) -1.
Given function: [tex]f(x) = \frac{4}{x} [/tex]
The derivative of f(x) with respect to x is:
[tex] \frac{df}{dx} = \frac{d}{dx}(4x^{-1}) [/tex]
[tex]\frac{df}{dx} = 4 * -1 * x^{-1 -1} \\ \frac{df}{dx} = -4 * x^{-2} [/tex]
At x = 2:
[tex]\frac{df}{dx} |_{x=2} = -4 * (2)^{-2} = -1[/tex]
Hence the correct answer is Option (B) -1.
(4) Ans: Option (C) 9 divided by 16.
Given function: [tex]f(x) = \frac{-9}{x} [/tex]
The derivative of f(x) with respect to x is:
[tex] \frac{df}{dx} = \frac{d}{dx}(-9x^{-1}) [/tex]
[tex]\frac{df}{dx} = -4 * -1 * x^{-1 -1} \\ \frac{df}{dx} = 9 * x^{-2} [/tex]
At x = -4:
[tex]\frac{df}{dx} |_{x=-4} = 9 * (-4)^{-2} = \frac{9}{16} [/tex]
Hence the correct answer is Option (C) 9 divided by 16.
(5) Ans: Option (D) 5
Given function:
[tex]f(x) = x^2 + 5 [/tex]
Now apply the limit:
[tex] \lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 + 5) \\ \lim_{x \to 0} f(x) = 0^2 + 5\\ \lim_{x \to 0} f(x) = 5[/tex]
The correct answer is Option (D) 5.
(6) Ans: Option (D) 27
Given function:
[tex]f(x) = x^2 + 3x - 1[/tex]
Apply the limit:
[tex] \lim_{x \to 4} f(x) = \lim_{x \to 4} (x^2 + 3x - 1) \\ \lim_{x \to 4} f(x) = (4)^2 + 3(4) -1 = 27[/tex]
The correct option is (D) 27
(7) Ans: Option (D) 8.
Given function:
[tex]f(x) = \frac{x^2 - 16}{x-4} [/tex]
[tex]f(x) = \frac{x^2 - 16}{x-4} \\ f(x) = \frac{(x+4)(x-4)}{(x-4)} \\ f(x) = x+4 [/tex]
Now apply limit:
[tex] \lim_{x \to 4} f(x) = \lim_{x \to 4} (x + 4) = 4+4 = 8[/tex]
The correct option is (D) 8.
(8) Ans: Option (A) Does not exist.
Given function:
[tex]f(x) = \frac{x^2 - 2x}{x^4} = \frac{x(x-2)}{x^4} = \frac{x-2}{x^3} [/tex]
Apply limit:
[tex] \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x-2}{x^3} = \frac{-2}{0^3} = -\infty [/tex]
The correct answer is (A) Does not exist.
(9),(10)
Please attach the graphs! Thanks! :)
(11) Ans: limit doesn't exist (Option C)
Given function:
[tex]f(x) = \left \{ {{x+1~~~~~x\ \textless \ -1} \atop {1-x~~~~~x\geq -1}} \right. [/tex]
If both sides are equal on applying limit then limit does exist.
Let check:
If x<-1: answer would be -1+1 = 0
If x≥-1: answer would be 1-(-1) =2
Since both are not equal, as 0≠2, hence limit doesn't exist (Option C).
(12) Ans: Option (B) 7.
Given function:
[tex]f(x) = \left \{ {{7-x^2~~~~~x\ \textless \ 0} \atop {-10x+7~~~~~x\textgreater 0}} \right. \\ and \\ f(x) = 7~~~~~~~x=0 [/tex]
If all of above three are equal upon applying limit, then limit exists.
When x < 0 -> 7-(0)^2 = 7
When x = 0 -> 7
When x > 0 -> -10(0) + 7 = 7
ALL of the THREE must be equal. As they are equal. Hence the correct option is (B) 7.
(13) Ans: -∞, x =3 (Option C)
Given function:
f(x) = 1/(x-3).
Table:
x f(x)=1/(x-3)
----------------------------------------
2.9 -10
2.99 -100
2.999 -1000
2.9999 -10000
3.0 -∞
Below the graph is attached! As you can see in the graph that at x=3, the curve approaches but NEVER exactly touches the x=3 line. Also the curve is in downward direction when you approach from the left. Hence, -∞, x =3 (Option C)
(14) Ans: Inst. velocity = -13
Given function:
s(t) = -1 -13t
Instantaneous velocity = [tex] \frac{ds(t)}{dt} [/tex]
Therefore,
[tex]\frac{ds(t)}{dt} = \frac{d}{dt} (-1-13t) = -13 [/tex]
At t=8:
Inst. velocity = -13
(15) Ans: +∞, x =1
Given function:
f(x) = 1/(x-1)^2
Table:
x f(x)= 1/(x-1)^2
----------------------------------
0.9 +100
0.99 +10000
0.999 +1000000
0.9999 +100000000
1.0 +∞
Below the graph is attached! As you can see in the graph that at x=1, the curve approaches but NEVER exactly touches the x=1 line. The curve is in upward direction if approached from left or right. Hence, the correct answer is: +∞, x =1
Given function: [tex]f(x) = -4x^2 + 11x[/tex]
The derivative of f(x) with respect to x is:
[tex] \frac{df}{dx} = -4 * 2x + 11 [/tex] --- (1)
Plug the value of x = 10 in (1)
(1) => [tex]\frac{df}{dx} |_{x=10} = -4 * 2(10) + 11 = -69[/tex]
Hence the correct answer is Option (B) -69
(2) Ans: Option (C) 8
Given function: [tex]f(x) = 8x + 4[/tex]
The derivative of f(x) with respect to x is:
[tex] \frac{df}{dx} = 8 [/tex] --- (1)
Plug the value of x = 9 in (1)
(1) => [tex]\frac{df}{dx} |_{x=9} = 8[/tex]
Hence the correct answer is Option (C) 8
(3) Ans: Option (B) -1.
Given function: [tex]f(x) = \frac{4}{x} [/tex]
The derivative of f(x) with respect to x is:
[tex] \frac{df}{dx} = \frac{d}{dx}(4x^{-1}) [/tex]
[tex]\frac{df}{dx} = 4 * -1 * x^{-1 -1} \\ \frac{df}{dx} = -4 * x^{-2} [/tex]
At x = 2:
[tex]\frac{df}{dx} |_{x=2} = -4 * (2)^{-2} = -1[/tex]
Hence the correct answer is Option (B) -1.
(4) Ans: Option (C) 9 divided by 16.
Given function: [tex]f(x) = \frac{-9}{x} [/tex]
The derivative of f(x) with respect to x is:
[tex] \frac{df}{dx} = \frac{d}{dx}(-9x^{-1}) [/tex]
[tex]\frac{df}{dx} = -4 * -1 * x^{-1 -1} \\ \frac{df}{dx} = 9 * x^{-2} [/tex]
At x = -4:
[tex]\frac{df}{dx} |_{x=-4} = 9 * (-4)^{-2} = \frac{9}{16} [/tex]
Hence the correct answer is Option (C) 9 divided by 16.
(5) Ans: Option (D) 5
Given function:
[tex]f(x) = x^2 + 5 [/tex]
Now apply the limit:
[tex] \lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 + 5) \\ \lim_{x \to 0} f(x) = 0^2 + 5\\ \lim_{x \to 0} f(x) = 5[/tex]
The correct answer is Option (D) 5.
(6) Ans: Option (D) 27
Given function:
[tex]f(x) = x^2 + 3x - 1[/tex]
Apply the limit:
[tex] \lim_{x \to 4} f(x) = \lim_{x \to 4} (x^2 + 3x - 1) \\ \lim_{x \to 4} f(x) = (4)^2 + 3(4) -1 = 27[/tex]
The correct option is (D) 27
(7) Ans: Option (D) 8.
Given function:
[tex]f(x) = \frac{x^2 - 16}{x-4} [/tex]
[tex]f(x) = \frac{x^2 - 16}{x-4} \\ f(x) = \frac{(x+4)(x-4)}{(x-4)} \\ f(x) = x+4 [/tex]
Now apply limit:
[tex] \lim_{x \to 4} f(x) = \lim_{x \to 4} (x + 4) = 4+4 = 8[/tex]
The correct option is (D) 8.
(8) Ans: Option (A) Does not exist.
Given function:
[tex]f(x) = \frac{x^2 - 2x}{x^4} = \frac{x(x-2)}{x^4} = \frac{x-2}{x^3} [/tex]
Apply limit:
[tex] \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x-2}{x^3} = \frac{-2}{0^3} = -\infty [/tex]
The correct answer is (A) Does not exist.
(9),(10)
Please attach the graphs! Thanks! :)
(11) Ans: limit doesn't exist (Option C)
Given function:
[tex]f(x) = \left \{ {{x+1~~~~~x\ \textless \ -1} \atop {1-x~~~~~x\geq -1}} \right. [/tex]
If both sides are equal on applying limit then limit does exist.
Let check:
If x<-1: answer would be -1+1 = 0
If x≥-1: answer would be 1-(-1) =2
Since both are not equal, as 0≠2, hence limit doesn't exist (Option C).
(12) Ans: Option (B) 7.
Given function:
[tex]f(x) = \left \{ {{7-x^2~~~~~x\ \textless \ 0} \atop {-10x+7~~~~~x\textgreater 0}} \right. \\ and \\ f(x) = 7~~~~~~~x=0 [/tex]
If all of above three are equal upon applying limit, then limit exists.
When x < 0 -> 7-(0)^2 = 7
When x = 0 -> 7
When x > 0 -> -10(0) + 7 = 7
ALL of the THREE must be equal. As they are equal. Hence the correct option is (B) 7.
(13) Ans: -∞, x =3 (Option C)
Given function:
f(x) = 1/(x-3).
Table:
x f(x)=1/(x-3)
----------------------------------------
2.9 -10
2.99 -100
2.999 -1000
2.9999 -10000
3.0 -∞
Below the graph is attached! As you can see in the graph that at x=3, the curve approaches but NEVER exactly touches the x=3 line. Also the curve is in downward direction when you approach from the left. Hence, -∞, x =3 (Option C)
(14) Ans: Inst. velocity = -13
Given function:
s(t) = -1 -13t
Instantaneous velocity = [tex] \frac{ds(t)}{dt} [/tex]
Therefore,
[tex]\frac{ds(t)}{dt} = \frac{d}{dt} (-1-13t) = -13 [/tex]
At t=8:
Inst. velocity = -13
(15) Ans: +∞, x =1
Given function:
f(x) = 1/(x-1)^2
Table:
x f(x)= 1/(x-1)^2
----------------------------------
0.9 +100
0.99 +10000
0.999 +1000000
0.9999 +100000000
1.0 +∞
Below the graph is attached! As you can see in the graph that at x=1, the curve approaches but NEVER exactly touches the x=1 line. The curve is in upward direction if approached from left or right. Hence, the correct answer is: +∞, x =1

