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Calculus 1 : Functions

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Example Questions

Example Question #2441 : Calculus

Find the general solution of the following differential equation:

$\frac{\mathrm{dy} }{\mathrm{d} x}=3y\sec^2(x)$

Possible Answers:

$y=Ce^{3\tan(x)}$

$y=Ce^{3\sec(x)tan(x)}$

y=x

$y=e^{3\tan(x)}+C$

Correct answer:

$y=Ce^{3\tan(x)}$

Explanation:

To find the general solution for the separable differential equation, we must move x and dx, y and dy to separate sides, and then integrate both sides:

$\int \frac{dy}{3y}=\int \sec^2(x)dx$

Next, integrate both sides:

$\frac{1}{3}\ln\left | y\right |=\tan(x)+C$

The rules used for the integrations are:

$\int \frac{dx}{x}=\ln\left | x\right |+C$ , $\int \sec^2(x)dx=\tan(x)+C$

Note that both Cs were combined to make one constant of integration in our equation.

Finally, solve for y:

$y=Ce^{3\tan(x)}$

Note that C was brought to the front, as e^C is itself a constant of integration.

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Example Question #111 : Solutions To Differential Equations

Derivative Rules

What is the derivative power rule?

Possible Answers:

For positive integers n, $\frac{d}{dx}x^{n}=nx^{n-1}$

For positive integers n, $\frac{d}{dx}x^{n}=(n-1)x^{n}$

For positive integers e, $\frac{d}{dx}e^{x}=e^x$

For all integers n, $\frac{d}{dx}x^{n}=nx^{n-1}$

For positive integers n, $\frac{d}{dx}x^{n}=nx^{1-n}$

Correct answer:

For positive integers n, $\frac{d}{dx}x^{n}=nx^{n-1}$

Explanation:

\frac{d}{dx} x^n = nx^{n-1} , \qquad n \neq 0.

The power rule holds for all powers except for the constant value $x^0$

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Example Question #1421 : Functions

Find f'(1) given:

f(x)=-3x^2-4x+8

Possible Answers:

Correct answer:

Explanation:

To solve, take the first derivative and evaluate at . Thus,

f'(x)=-6x-4

f'(1)=-6(1)-4=-10

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Example Question #1422 : Functions

Solve the initial value problem: y' +3y = 1, y(0) =1 .

Possible Answers:

$y = \frac{1}{3}+\frac{1}{2e^{3t}}$

$y = \frac{1}{3}+\frac{2}{3e^{3t}}$

$y = \frac{1}{3}+\frac{1}{3e^{3t}}$

$y = \frac{2}{3}+\frac{2}{3e^{3t}}$

$y = \frac{1}{3}+\frac{3}{2e^{3t}}$

Correct answer:

$y = \frac{1}{3}+\frac{2}{3e^{3t}}$

Explanation:

The differential equation is in its correct form.

$\frac{dy}{dx} +p(t)y = g(t)$

Solve for the integrating factor.

$u(t) = e^{\int p(t)dt}$

$u(t) = e^{\int 3dt} = e^{3t}$

Multiply the integration factor throughout the entire equation.

$e^{3t}[y' +3y = 1]$

$e^{3t}y' +3e^{3t}y = e^{3t}$

The left side of the equation becomes $(e^{3t}y)'$ from the use of our integrating factor. Rewrite the equation.

$(e^{3t}y)'=e^{3t}$

Integrate both sides.

$\int (e^{3t}y)' \:dt=\int e^{3t}\:dt$

$e^{3t}y+C_1= \frac{1}{3}e^{3t}+C_2$

Merge the constants.

$e^{3t}y = \frac{1}{3}e^{3t}+C$

Divide by $e^{3t}$ on both sides.

$y = \frac{1}{3}+\frac{C}{e^{3t}}$

Substitute the initial condition y(0) =1 to solve for .

$1 = \frac{1}{3}+\frac{C}{e^{3(0)}}$

$\frac{2}{3}=\frac{C}{e^{3(0)}}$

$\frac{2e^{3(0)}}{3}=C$

$C=\frac{2}{3}$

Resubstitute the constant. The answer is:

$y = \frac{1}{3}+\frac{2}{3e^{3t}}$

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Example Question #1422 : Functions

Find f'(-1) given:

f(x)=x^2-16x+11

Possible Answers:

Correct answer:

Explanation:

To solve, simply find the first derivative and let x=-1 . Thus,

$f'(x)=2x-16\Rightarrow f'(-1)=2(-1)-16=-2-16=-18$

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Example Question #1423 : Functions

In which interval(s) is f(x) = x^3 -3x^2 -9x increasing?

Possible Answers:

$(3, \infty)$

$(1, \infty)$

$(-\infty, -1) \cup (3, \infty)$

(-1,3)

Correct answer:

$(-\infty, -1) \cup (3, \infty)$

Explanation:

f(x) is increasing when f'(x) > 0 .

$f'(x) = 3x^2-6x-9 > 0 \Rightarrow 3(x^2-2x-3) > 0 \Rightarrow x^2-2x-3 > 0 \\ \Rightarrow (x-3)(x+1)>0 \Rightarrow x >3 \or \ x<-1. (x \in (-\infty, -1) \cup (3, \infty)).$

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Example Question #122 : How To Find Solutions To Differential Equations

Find f'(3) given:

$f(x)=\ln(x)+\frac{1}{9}x^2+5x$

Possible Answers:

Correct answer:

Explanation:

To solve, simply differentiate and plug in x=3 . Thus,

$f'(t)=\frac{1}{x}+\frac{2}{9}x+5$

$f'(3)=\frac{1}{3}+\frac{2}{3}+5=6$

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Example Question #373 : Equations

Find the general solution to the following differential equation:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{e^x+\sin(x)}{y}$

Possible Answers:

$y=e^x+\cos(x)+C$

$y=\sqrt{2e^x-2\cos(x)}+C$

$y=\sqrt{e^x-\cos(x)}+C$

$y=\sqrt{2e^x+2\cos(x)}+C$

$y=\ln\left | y\right |\cdot (e^x-\cos(x))+C$

Correct answer:

$y=\sqrt{2e^x-2\cos(x)}+C$

Explanation:

To solve the separable differential equation, we must separate the x and dx, y and dy terms, to opposite sides of the equation:

$y dy=(e^x+\sin(x))dx$

Now, integrate both sides to solve for y:

$\frac{1}{2}y^2=e^x-\cos(x)+C$

The following rules were used for integration:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int e^xdx=e^x+C$ , $\int \sin(x)dx=-\cos(x)+C$

Note that we combined the C's to make one constant of integration.

Finally, isolate y by itself:

$y=\sqrt{2e^x-2\cos(x)}+C$

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Example Question #133 : Differential Equations

If $f(x) = 7x^{4} + 5x^{3} + 6x^{2} + 7x + 9$ , what does f'(x) equal?

Possible Answers:

$f'(x) = 21x^{3} + 23x^{2} + 5x$

$f'(x) = 28x^{5} + 15x^{4} + 12x^{3} + 14x^{2} + 9x$

$f'(x) = 21x^{3} + 10x^{2} + 6x + 7$

$f'(x) = 28x^{4} + 15x^{3} + 12x^{2} + 7x$

$f'(x) = 28x^{3} + 15x^{2} + 12x + 7$

Correct answer:

$f'(x) = 28x^{3} + 15x^{2} + 12x + 7$

Explanation:

In order to find the derivative of a function using the Power Rule, mulitply the constant in front of the variable (x) by the current power that the variable is raised to, then decrease that power by 1. For example, the derivative of Axⁿis equal to (n*A)x^n-1. If there is no variable, and just a constant, then that constant will not be changing over time, and is the derivative (rate of change) will be equal to zero.

$f(x) = 7x^{4} + 5x^{3} + 6x^{2} + 7x + 9$

$f'(x) = (7*4)x^{4-1}+(5*3)x^{3-1}+(6*2)x^{2-1}+(7*1)x^{1-1}$

$f'(x) = 28x^{3} + 15x^{2} + 12x + 7$

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Example Question #375 : Equations

A function f(x) is given by the formula

$f(x) = \frac{1}{3}x^3 - \frac{3}{4}x^2 -5x+2$ .

Which values corresponds to the local maximum and local minimum of this function?

Possible Answers:

$x_{max} = 3.1, x_{min} =-1.6$

$x_{max} = -2.6, x_{min} = 1.6$

$x_{max} = -4.3, x_{min} = 2.9$

$x_{max} = -1.6, x_{min} = 1.6$

$x_{max} = -1.6, x_{min} = 3.1$

Correct answer:

$x_{max} = -1.6, x_{min} = 3.1$

Explanation:

In order to find the local maximum, the derivate of the function f(x) must be taken. This is because the derivative of f(x) gives the rate of change for f(x). When f(x) is approaching its maximum, it is increasing. When it reaches the maximum, it stays constant (at the very top before it being to decrease, there is a point where the slope of the tangent line is equal to zero). Directly after reaching the maximum, f(x) begins to decrease. Therefore, the local maximum can be found using the derivative and noting where f'(x) crosses the x-axis, known as the x-intercept. This point could be either a local maximum of minimum, depending on the slope of f'(x). If f'(x) is decreasing, it is approaching a local maximum, and if it is increasing, a local minimum.

$f'(x) = x^{2}-1.5x-5$

Setting this equal to zero and solving using the quadratic formula:

$\frac{1.5\pm \sqrt{1.5^{2}-4*1*-5} }{2}$

This gives us x = -1.6 or x = 3.1. In order to determine if these indicate a local maximum of minimum, the second derivative will show if the slope is positive or negative at these points. If f''(x) is negative, it is a local maximum, if positive, it is a local minimum.

f''(x) = 2x - 1.5

f''(-1.6) = -4.7

f''(3.1) = 4.7

Therefore, the local maximum is located at x = -1.6.

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2441 : Calculus

Example Question #111 : Solutions To Differential Equations

Example Question #1421 : Functions

Example Question #1422 : Functions

Example Question #1422 : Functions

Example Question #1423 : Functions

Example Question #122 : How To Find Solutions To Differential Equations

Example Question #373 : Equations

Example Question #133 : Differential Equations

Example Question #375 : Equations

All Calculus 1 Resources

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