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Calculus 1 : Writing Equations

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Example Questions

Example Question #2291 : Calculus

Find the equation of the line tangent to $f(x)=3x^{3}-x^{2}+1$ at x=-2 .

Possible Answers:

y=-40x+53

y=20x-17

The tangent line does not exist at this point.

y=x+53

y=40x+53

Correct answer:

y=40x+53

Explanation:

To find the equation for the tangent line, we must first find the derivative of f(x) .

In order to evaluate this derivative, we must use these formulae:

$\frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1, \frac{d}{dx}cx=c$

$\frac{d}{dx}(x^{n}))=nx^{n-1}, \frac{d}{dx}(cx^{n})=ncx^{n-1}$

$f'(x)=3*3x^{3-1}-2x^{2-1}+0=9x^{2}-2x$

Next we find the slope of the tangent line at this point:

$f'(-2)=9(-2)^{2}-2(-2)=36+4=40$

Then we need to plug the given value back into the original function in order to find the corresponding value.

$f(-2)=3*(-2)^{3}-(-2)^{2}+1=-24-4+1=-27$

Now that we have a point (-2,-27) and a slope of , we can find the equation for the line by plugging these values into y=mx+b form to solve for :

-27=40*-2+b ==> b = 53

So, the equation for the line is y=40x+53 .

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Example Question #21 : How To Write Equations

Give the position equation p(t) for an object falling towards Earth with a constant acceleration of $-9.8 \frac{m}{s^2}$ , with initial velocity of $13 \frac{m}{s}$ , and initial position at 35m above the ground.

Possible Answers:

p(t)=-4.9t^2+13t+35

p(t)=-9.8t^2+13t+35

p(t)=-4.9t^2+35t+13

p(t)=4.9t^2+13t+35

Correct answer:

p(t)=-4.9t^2+13t+35

Explanation:

For this equation, let's start with acceleration and move backwards towards position. We know that the object is accelerating at a constant rate of -9.8

p''(t)=-9.8

To solve for velocity , we integrate both sides with respect to

$\int p''(t)dt=\int -9.8 dt$

p'(t)=-9.8t+B , where is a constant

Since we know that the initial velocity is , p'(0)=13, B=13 .

p'(t)=-9.8t+13

To solve for position we have to take the integral of both sides with respect to .

$\int p'(t)dt=\int (-9.8t+13)dt$

p(t)=-4.9t^2+13t+C , where is a constant.

Since we know that the initial position is , P(0)=35 , C=35

The formula for this motion can be modelled by:

p(t)=-4.9t^2+13t+35

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Example Question #1261 : Functions

Inverse Function

What is the inverse function of the following:

4y-x=4

Possible Answers:

4x=y-4

None of the above

$x-{\frac14}y=1$

4x-y=1

4x+y+4=0

Correct answer:

$x-{\frac14}y=1$

Explanation:

4y-x=4 can be written as $y=\frac{1}{4}x+1$ with slope of $\frac{1}{4}$ and y-intercept of 1. So the inverse function would be $x=\frac{1}{4}y+1$ or, $x-{\frac14}y=1$

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Example Question #2294 : Calculus

Logarithm Functions

Solve for ,

$\ln x=-9$

Possible Answers:

$-e\sqrt{9}$

e-9

$e^{-9}$

9-e

Correct answer:

$e^{-9}$

Explanation:

$\ln x=-9$ implies $e^{lnx}=e^{-9}$ implies $x=e^{-9}$

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Example Question #21 : How To Write Equations

Write the integral expression for how to find the area under the curve f(x) from the point (-3,4) to (5,6) .

f(x)=4x^3-2x+7

Possible Answers:

$\int_{-3}^{6}(4x^3-2x+7)dx$

$\int_{4}^{6}(4x^3-2x+7)dx$

$\int_{-3}^{5}(4x^3-2x+7)dx$

$\int_{4}^{5}(4x^3-2x+7)dx$

Correct answer:

$\int_{-3}^{5}(4x^3-2x+7)dx$

Explanation:

To write the expression, simply use the bounds laid out by the x-coordinates of the points given. Thus, your answer is:

$\int_{-3}^{5}(4x^3-2x+7)dx$

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Example Question #21 : How To Write Equations

$f(x) = \left\{\begin{matrix} cx^2-2x & x<3\\ x& x \geq 3 \end{matrix}\right.$

For what values of the constant is the function f(x) continuous on $(-\infty, \infty)$ ?

Possible Answers:

Correct answer:

Explanation:

Both cx^2-2x and are continuous in the intervals they are defined. In order for the function to be continuous on the entire real line, we need to have cx^2 -2x =x at x =3 . Plugging the value in, we get $c \cdot 3^2 - 2 \cdot 3 = 3$ $\Rightarrow 9c-6 =3 \Rightarrow c=1.$

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Example Question #22 : How To Write Equations

White the expression for the derivative of the following function.

p(t)=2t^2-1

Possible Answers:

$p'(t)=\frac{2}{3}t^3-t$

p'(t)=2t^2-1

p'(t)=4t

p(t)=4t

Correct answer:

p'(t)=4t

Explanation:

To solve, simply differentiate and don't forget to indicate the differentiation by the "apostrophe" between the function and dependent variable. Thus,

p'(t)=4t

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Example Question #23 : How To Write Equations

Find the equation for the velocity whose position function is below:

p(t)=5t^2-6t-3

Possible Answers:

v(t)=10

v(t)=10t+6

v(t)=10t-6

v(t)=5t^2-6t-3

Correct answer:

v(t)=10t-6

Explanation:

To solve, simply differentiate p(t) .

Remember to use the power rule.

Recall the power rule:

f(x)=x^n

$f'(x)=nx^{n-1}$

Apply this to our situation to get

v(t)=p'(t)=10t-6

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Example Question #31 : Other Writing Equations

Find the second derivative of the following function:
$g(x)=e^{2x}-Sin(x)+40$

Possible Answers:

$g''(x)=4e^{x}+Sin(x)$

$g''(x)=4e^{2x}+Sin(x)$

$g''(x)=4e^{4x}-Sin(x)$

$g''(x)=2e^{x}-Cos(x)$

$g''(x)=2e^{2x}-Cos(x)$

Correct answer:

$g''(x)=4e^{2x}+Sin(x)$

Explanation:

The second derivative is the derivative of the first derivative function.

With our original function, $g(x)=e^{2x}-Sin(x)+40$

We may find the first derivative to be: $g'(x)=2e^{2x}-Cos(x)$

Note that the chain rule was used to find the first derivative.

Now to find the second derivative we must take the derivative of this function:

$g''(x)=4e^{2x}+Sin(x)$

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Example Question #31 : How To Write Equations

You wish to find the area under the line y= 2x + 5 between the values of x=3 and x=100 . Which is the correct way to express this task as an equation?

Possible Answers:

$\int_{3}^{100}(x^{2}+5x+C)dx$

$\int_{3}^{100}(2x+5)$

$\int_{3}^{100}(x^{2}+5x)dx$

$\int_{3}^{100}(2x+5)dx$

$\int_{3}^{100}(x^{2}+5x+C)$

Correct answer:

$\int_{3}^{100}(2x+5)dx$

Explanation:

To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 3. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 100. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.

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Calculus 1 : Writing Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2291 : Calculus

Example Question #21 : How To Write Equations

Example Question #1261 : Functions

Example Question #2294 : Calculus

Example Question #21 : How To Write Equations

Example Question #21 : How To Write Equations

Example Question #22 : How To Write Equations

Example Question #23 : How To Write Equations

Example Question #31 : Other Writing Equations

Example Question #31 : How To Write Equations

All Calculus 1 Resources

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