Hello Sarah! I'd be happy to generate 10 calculus questions with a medium difficulty level. Here are your questions:
**Question 1**
Find the derivative of the function `f(x) = 3x^2 + 2x - 5` using the limit definition of a derivative.
[Image: A graph of the function f(x) with a tangent line at x = 1]
**Answer**
Using the limit definition of a derivative, we have:
```
f'(x) = lim(h → 0) [f(x + h) - f(x)]/h
= lim(h → 0) [(3(x + h)^2 + 2(x + h) - 5) - (3x^2 + 2x - 5)]/h
= lim(h → 0) [3x^2 + 6xh + 3h^2 + 2x + 2h - 5 - 3x^2 - 2x + 5]/h
= lim(h → 0) [6xh + 3h^2 + 2h]/h
= 6x + 3h + 2
= 6x + 2 (since h → 0)
```
So, `f'(x) = 6x + 2`.
**Question 2**
Evaluate the following definite integral: `∫(x^2 + 1) dx from x = 0 to x = 2`.
[Image: A graph of the function x^2 + 1 with the area shaded between x = 0 and x = 2]
**Answer**
Using the power rule of integration, we have:
```
∫(x^2 + 1) dx = (1/3)x^3 + x | from x = 0 to x = 2
= [(1/3)(2^3) + 2] - [(1/3)(0^3) + 0]
= (8/3) + 2
= 14/3
```
So, the definite integral is `14/3`.
**Question 3**
Find the area under the curve `y = x^2` from `x = 1` to `x = 4`.
[Image: A graph of the function y = x^2 with the area shaded between x = 1 and x = 4]
**Answer**
Using the definite integral, we have:
```
Area = ∫(x^2) dx from x = 1 to x = 4
= (1/3)x^3 | from x = 1 to x = 4
= [(1/3)(4^3) - (1/3)(1^3)]
= (64/3) - (1/3)
= 63/3
= 21
```
So, the area under the curve is `21`.
**Question 4**
A particle moves along a straight line with velocity `v(t) = 2t^2 + 3t - 4` meters per second. Find the acceleration of the particle at `t = 2` seconds.
[Image: A graph of the velocity function v(t) with a tangent line at t = 2]
**Answer**
Using the definition of acceleration, we have:
```
a(t) = dv/dt
= d(2t^2 + 3t - 4)/dt
= 4t + 3
```
At `t = 2` seconds, the acceleration is:
```
a(2) = 4(2) + 3
= 11
```
So, the acceleration of the particle at `t = 2` seconds is `11` meters per second squared.
**Question 5**
Find the value of `x` that satisfies the equation `∫(2x) dx from x = 1 to x = x = 4`.
[Image: A graph of the function 2x with the area shaded between x = 1 and x = x]
**Answer**
Using the definite integral, we have:
```
∫(2x) dx from x = 1 to x = x
= x^2 | from x = 1 to x = x
= x^2 - 1
```
Since the area is equal to `4`, we set up the equation:
```
x^2 - 1 = 4
x^2 = 5
x = ±√5
```
So, the values of `x` that satisfy the equation are `x = √5` and `x = -√5`.
**Question 6**
Find the derivative of the function `f(x) = sin(x)/x` using the quotient rule.
[Image: A graph of the function f(x) with a tangent line at x = π/4]
**Answer**
Using the quotient rule, we have:
```
f'(x) = (cos(x) - sin(x))/x^2
```
So, the derivative of the function is `(cos(x) - sin(x))/x^2`.
**Question 7**
Evaluate the following definite integral: `∫(e^x) dx from x = 0 to x = 1`.
[Image: A graph of the function e^x with the area shaded between x = 0 and x = 1]
**Answer**
Using the exponential rule of integration, we have:
```
∫(e^x) dx from x = 0 to x = 1
= e^x | from x = 0 to x = 1
= e - 1
```
So, the definite integral is `e - 1`.
**Question 8**
Find the area under the curve `y = x^3` from `x = 0` to `x = 1`.
[Image: A graph of the function y = x^3 with the area shaded between x = 0 and x = 1]
**Answer**
Using the definite integral, we have:
```
Area = ∫(x^3) dx from x = 0 to x = 1
= (1/4)x^4 | from x = 0 to x = 1
= (1/4)(1^4) - (1/4)(0^4)
= 1/4
```
So, the area under the curve is `1/4`.
**Question 9**
A particle moves along a curve with position `s(t) = t^3 - 2t^2 + t + 1` meters at time `t` seconds. Find the acceleration of the particle at `t = 1` second.
[Image: A graph of the position function s(t) with a tangent line at t = 1]
**Answer**
Using the definition of acceleration, we have:
```
a(t) = d^2s/dt^2
= d(t^3 - 2t^2 + t + 1)/dt^2
= 6t - 4
```
At `t = 1` second, the acceleration is:
```
a(1) = 6(1) - 4
= 2
```
So, the acceleration of the particle at `t = 1` second is `2` meters per second squared.
**Question 10**
Find the value of `x` that satisfies the equation `∫(x^2 + 1) dx from x = 0 to x = x = 2`.
[Image: A graph of the function x^2 + 1 with the area shaded between x = 0 and x = x]
**Answer**
Using the definite integral, we have:
```
∫(x^2 + 1) dx from x = 0 to x = x
= (1/3)x^3 + x | from x = 0 to x = x
= (1/3)x^3 + x - 0
```
Since the area is equal to `2`, we set up the equation:
```
(1/3)x^3 + x = 2
x^3 + 3x - 6 = 0
(x - 1)(x^2 + x + 6) = 0
x = 1
```
So, the value of `x` that satisfies the equation is `x = 1`.
I hope you find these questions helpful, Sarah! Let me know if you need anything else.
Notis
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Craft custom math challenges with instant solutions.OpenHello Sarah! I'd be happy to generate 10 calculus questions with a medium difficulty level. Here are your questions: **Question 1** Find the derivative of the function `f(x) = 3x^2 + 2x - 5` using the limit definition of a derivative. [Image: A graph of the function f(x) with a tangent line at x = 1] **Answer** Using the limit definition of a derivative, we have: ``` f'(x) = lim(h → 0) [f(x + h) - f(x)]/h = lim(h → 0) [(3(x + h)^2 + 2(x + h) - 5) - (3x^2 + 2x - 5)]/h = lim(h → 0) [3x^2 + 6xh + 3h^2 + 2x + 2h - 5 - 3x^2 - 2x + 5]/h = lim(h → 0) [6xh + 3h^2 + 2h]/h = 6x + 3h + 2 = 6x + 2 (since h → 0) ``` So, `f'(x) = 6x + 2`. **Question 2** Evaluate the following definite integral: `∫(x^2 + 1) dx from x = 0 to x = 2`. [Image: A graph of the function x^2 + 1 with the area shaded between x = 0 and x = 2] **Answer** Using the power rule of integration, we have: ``` ∫(x^2 + 1) dx = (1/3)x^3 + x | from x = 0 to x = 2 = [(1/3)(2^3) + 2] - [(1/3)(0^3) + 0] = (8/3) + 2 = 14/3 ``` So, the definite integral is `14/3`. **Question 3** Find the area under the curve `y = x^2` from `x = 1` to `x = 4`. [Image: A graph of the function y = x^2 with the area shaded between x = 1 and x = 4] **Answer** Using the definite integral, we have: ``` Area = ∫(x^2) dx from x = 1 to x = 4 = (1/3)x^3 | from x = 1 to x = 4 = [(1/3)(4^3) - (1/3)(1^3)] = (64/3) - (1/3) = 63/3 = 21 ``` So, the area under the curve is `21`. **Question 4** A particle moves along a straight line with velocity `v(t) = 2t^2 + 3t - 4` meters per second. Find the acceleration of the particle at `t = 2` seconds. [Image: A graph of the velocity function v(t) with a tangent line at t = 2] **Answer** Using the definition of acceleration, we have: ``` a(t) = dv/dt = d(2t^2 + 3t - 4)/dt = 4t + 3 ``` At `t = 2` seconds, the acceleration is: ``` a(2) = 4(2) + 3 = 11 ``` So, the acceleration of the particle at `t = 2` seconds is `11` meters per second squared. **Question 5** Find the value of `x` that satisfies the equation `∫(2x) dx from x = 1 to x = x = 4`. [Image: A graph of the function 2x with the area shaded between x = 1 and x = x] **Answer** Using the definite integral, we have: ``` ∫(2x) dx from x = 1 to x = x = x^2 | from x = 1 to x = x = x^2 - 1 ``` Since the area is equal to `4`, we set up the equation: ``` x^2 - 1 = 4 x^2 = 5 x = ±√5 ``` So, the values of `x` that satisfy the equation are `x = √5` and `x = -√5`. **Question 6** Find the derivative of the function `f(x) = sin(x)/x` using the quotient rule. [Image: A graph of the function f(x) with a tangent line at x = π/4] **Answer** Using the quotient rule, we have: ``` f'(x) = (cos(x) - sin(x))/x^2 ``` So, the derivative of the function is `(cos(x) - sin(x))/x^2`. **Question 7** Evaluate the following definite integral: `∫(e^x) dx from x = 0 to x = 1`. [Image: A graph of the function e^x with the area shaded between x = 0 and x = 1] **Answer** Using the exponential rule of integration, we have: ``` ∫(e^x) dx from x = 0 to x = 1 = e^x | from x = 0 to x = 1 = e - 1 ``` So, the definite integral is `e - 1`. **Question 8** Find the area under the curve `y = x^3` from `x = 0` to `x = 1`. [Image: A graph of the function y = x^3 with the area shaded between x = 0 and x = 1] **Answer** Using the definite integral, we have: ``` Area = ∫(x^3) dx from x = 0 to x = 1 = (1/4)x^4 | from x = 0 to x = 1 = (1/4)(1^4) - (1/4)(0^4) = 1/4 ``` So, the area under the curve is `1/4`. **Question 9** A particle moves along a curve with position `s(t) = t^3 - 2t^2 + t + 1` meters at time `t` seconds. Find the acceleration of the particle at `t = 1` second. [Image: A graph of the position function s(t) with a tangent line at t = 1] **Answer** Using the definition of acceleration, we have: ``` a(t) = d^2s/dt^2 = d(t^3 - 2t^2 + t + 1)/dt^2 = 6t - 4 ``` At `t = 1` second, the acceleration is: ``` a(1) = 6(1) - 4 = 2 ``` So, the acceleration of the particle at `t = 1` second is `2` meters per second squared. **Question 10** Find the value of `x` that satisfies the equation `∫(x^2 + 1) dx from x = 0 to x = x = 2`. [Image: A graph of the function x^2 + 1 with the area shaded between x = 0 and x = x] **Answer** Using the definite integral, we have: ``` ∫(x^2 + 1) dx from x = 0 to x = x = (1/3)x^3 + x | from x = 0 to x = x = (1/3)x^3 + x - 0 ``` Since the area is equal to `2`, we set up the equation: ``` (1/3)x^3 + x = 2 x^3 + 3x - 6 = 0 (x - 1)(x^2 + x + 6) = 0 x = 1 ``` So, the value of `x` that satisfies the equation is `x = 1`. I hope you find these questions helpful, Sarah! Let me know if you need anything else.
