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Calculus AB : Calculus AB

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Example Questions

Example Question #12 : Determine Local Linearity And Linearization

Find the equation of the line perpendicular to the line tangent to the following function at x=1, and passing through (0, 6):

f=x^3+3x

Possible Answers:

y=x+1

$y=\frac{-x}{6}+6$

y=x

y=6x+6

Correct answer:

$y=\frac{-x}{6}+6$

Explanation:

To find the equation of the line perpendicular to the tangent line to the function at a certain point, we must find the slope of the tangent line to the function, which is the derivative of the function at that point:

f'=3x^2+3

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}, \frac{\mathrm{d} }{\mathrm{d} x}ax=a$

Now, we evaluate the derivative at the given point:

3(1)+3=6=m

We now know the slope of the tangent line, but because the line we are solving for is perpendicular to this line, its slope is the negative reciprocal, $-\frac{1}{6}$ .

With a slope and a point, we can now find the equation of the line:

$y-6=-\frac{1}{6}(x+0)$

$y=\frac{-x}{6}+6$

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Example Question #13 : Determine Local Linearity And Linearization

Find the equation of the line that is tangent to the graph of $f(x) = sin(2\pi x) + x^{3}$ when x = 3 .

Possible Answers:

y-27 = 27(x-3)

$y+27 = (2\pi + 27)(x+3)$

$y-27 = (2\pi + 27)(x-3)$

$y-27 = 2\pi(x-3)$

y+27 = 27(x+3)

Correct answer:

$y-27 = (2\pi + 27)(x-3)$

Explanation:

First, evaluate $f(3)=sin(6\pi ) + 3^{3} = 27$ .

Then, to find the slope of the tangent line, find f'(3) .

$f'(x) = 2\pi cos(2\pi x) + 3x^{2}$ , so $f'(3) = 2\pi + 27$ .

Therefore, the equation of the tangent line is

$y-27 = (2\pi + 27)(x-3)$ .

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Example Question #364 : Calculus Ab

Find the minimum value of $f(x)=\frac{1}{\sqrt{9-x^2}}$

Possible Answers:

$\frac{1}{3}$

$-\frac{1}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

In order to find the extreme value, we need to take the derivative of the function.

$f'(x)=\frac{x}{(9-x^2)^{\frac{3}{2}}}$

After setting it equal to 0, we see that the only candidate is for x=0 . After setting x=0 into f(x) , we get the coordinate $(0,\frac{1}{3})$ as an extreme value. To confirm it is a minimum we can plot the function.

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Example Question #51 : Contextual Applications Of Derivatives

Let f(x)=x^4+2x^3+5x +1 . Find the equation of the line tangent to f(x) at the point (1,9) .

Possible Answers:

y = 15x-24

y = 15x-6

y = 12x -3

y=6

Correct answer:

y = 15x-6

Explanation:

To find the equation of a tangent line, we need two things: The tangent point, which is given as (1,9) , and the slope of the tangent line at that point, which is the derivative at that point.

To find the derivative at the point, we will find f'(x) , using derivative rules. Then we will plug the given point's x-coordinate into f'(x) and that will give us the slope we need.

Finding the derivative of f(x)=x^4+2x^3+5x +1 will require the power rule for each term. Recall that the power rule is $\frac{\mathrm{d} }{\mathrm{d} x}[x^n]= n x^{n-1}$ . For the , the power rule effectively removes the . Also, the derivative of a constant is , so the will drop when we get the derivative.

Applying these rules, we get

$f'(x)=(4)x^{4-1} + (3)2x^{3-1}+5 +0$

f'(x)=4x^3+6x^2 + 5

Now that we have the derivative, we effectively have a formula to find the slope of a tangent line at any point we choose. The question asks for the tangent line at (1,9) . So we plug the x-coordinate of the point into the derivative function to find the slope.

f'(1)=4(1)^3 + 6(1)^2 +5

f'(1)=4 + 6 + 5

f'(1)=15

The last step is to use the point-slope equation of a line to construct the equation we need. The point-slope equation is (y - y_1)= m(x-x_1) , where is the slope, and (x_1,y_1) is the given point.

Plugging our slope into , and our original point (1,9) in for x_1 and y_1 , we get

(y-9)= 15(x-1)

Now we can solve for to compare to the answer choices.

y-9=15x - 15

y = 15x - 15 +9

y = 15x - 6

This is the equation of the tangent line at the given point. We have an answer.

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Example Question #14 : Determine Local Linearity And Linearization

Let $h(x)=\sin(x)-2x\cos(x)$ . Find the equation of the line tangent to h(x) at the point $(\pi, 2\pi)$ .

Possible Answers:

$y = x + 3\pi$

$y=x+\pi$

$y = -3x + 5\pi$

y = x

Correct answer:

$y=x+\pi$

Explanation:

To find the equation of a tangent line at a point, we will need the slope of the function at that point. To find this, we find the derivative of the function.

Finding this derivative will use trigonometric function derivative rules, and the product rule.

Recall that the derivative of $\sin(x)$ is $\cos(x)$ . This takes care of the first term.

To find the derivative of the next term, we need to be careful with our signs. We will use the product rule which results in two terms. The negative sign can mess us up if we aren't careful. The way we will handle this is to associate the minus sign with the term when doing the product rule. So we will find the derivative of $-2x\cos(x)$ . The product rule is as follows $\frac{\mathrm{d} }{\mathrm{d} x}[{\color{Red} f(x)}\cdot {\color{Blue} g(x)}]= {\color{Red} f'(x)}{\color{Blue} g(x)}+{\color{Blue} g'(x)}{\color{Red} f(x)}$ , where the red and blue are the two factors of the term we are differentiating. In our case, ${\color{Red} -2x}$ and ${\color{Blue} \cos(x)}$ .

${\color{Red} f'(x)}= {\color{Red} -2}$ by the special case of the power rule. The drops.

${\color{Blue} g'(x)}={\color{Blue} -\sin(x)}$ by the trigonometric derivative rules.

Putting these together we get the derivative of $-2x\cos(x)$ to be

${\color{Red} -2}\cdot{\color{Blue} \cos(x)}+ {\color{Blue} (-\sin(x))}{\color{Red} (-2x)}$

Simplifying, we get

$-2\cos(x)+2x\sin(x)$

Assembling all the pieces of the derivative together, we have found

$h'(x)=\cos(x)-2\cos(x)+2x\sin(x)$

Combining like terms gives

$h'(x)=-\cos(x)+2x\sin(x)$

Now we have a formula for the tangent slope of h(x) at any point. The point we care about is $(\pi,2\pi)$ . To find this point's tangent slope, we will plug its x-coordinate into our derivative.

$h'(\pi)=-\cos(\pi)+2(\pi)\sin(\pi)$

Simplifying gives

$-(-1)+2(\pi)(0)$

So we have found the slope of the tangent line at our point, $(\pi,2\pi)$ , is .

The last step is the point-slope equation of a line, (y - y_1)=m(x-x_1) , where is the slope and (x_1,y_1) is the given point. Plugging in for , $\pi$ for x_1 , and $2\pi$ for y_1 , we get

$(y - 2\pi)= 1(x-\pi)$

Solving for , we get

$y - 2\pi=x - \pi$

$y = x -\pi + 2\pi$

$y = x + \pi$

This is the equation of the tangent line at the given point. We have an answer.

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Example Question #1 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Find the area of the region enclosed by the parabola $y=12-x^2$ and $y=-x$ .

Possible Answers:

$\frac{301}{6}$

$\frac{401}{6}$

$\frac{434}{6}$

$\frac{343}{6}$

$\frac{334}{6}$

Correct answer:

$\frac{343}{6}$

Explanation:

The two curves intersect in between x=-3 and x=4 , which can be found by solving the quadratic equation $12-x^2=-x$ .

To solve for the area between curves, y=f(x) and y=g(x) , we use the formula $A=\int^b_a(f(x)-g(x))dx$

For our problem:

$A=\int^4_{-3}(12-x^2-(-x))dx$

$A=12x-\frac{x^3}{3}+\frac{x^2}{2}$ evaluated from to which yields $\frac{343}{6}$ .

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Example Question #2 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Find the area under the curve $f(x)=\frac{1}{\sqrt{x+2}}$ between $2\leq x\leq 7$ .

Possible Answers:

$3$

$5$

$2$

$1$

$4$

Correct answer:

$2$

Explanation:

To find the area under the curve, we need to integrate. In this case, it is a definite integral.

$\int_{2}^{7}\frac{1}{\sqrt{x+2}}dx=2\sqrt{x+2}\Big|_2^7=2$

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Example Question #3 : Analytical Applications Of Derivatives

Find the area bounded by $y=2, y=x, y=\frac{1}{9}x^2, x=3$

Possible Answers:

2.5

1.5

3.5

Correct answer:

Explanation:

The easiest way to look at this is to plot the graphs. The shaded area is the actual area that we want to compute. We can first find area bounded by x=3 and y=2 in the first quadrant and subtract the excessive areas. The area of that rectangle box is 6. The area under the curve $\frac{1}{9}x^{2}$ is $\int_{0}^{3}\frac{1}{9}x^2=1$ .

The area of the triangle above the curve y=x is 2. Therefore, the area bounded is 6-1-2=3 .

Screen shot 2020 09 04 at 1.20.38 pm

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Example Question #1 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

$\int _0^\pi sin(x) dx$

Possible Answers:

Correct answer:

Explanation:

$\int sin(x) dx = -cos(x) + C$ $\int _0^\pi sin(x) dx = -cos(\pi) - -cos(0)= 1-(-1) = 2$

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Example Question #5 : Analytical Applications Of Derivatives

Consider the region bounded by the functions

$f(x) = -x^{2} + 2x$ and

$g(x) = -sin(\frac{\pi x}{2})$

between x = 0 and x = 2 . What is the area of this region?

Possible Answers:

$\frac{4}{3}$

$\frac{3\pi }{4}$

$\frac{3+\pi }{4}$

$\frac{4}{3} +\frac{4}{\pi }$

$\frac{\pi }{4}$

Correct answer:

$\frac{4}{3} +\frac{4}{\pi }$

Explanation:

The area of this region is given by the following integral:

$Area = \int_{0}^{2} f(x) - g(x) = \int_{0}^{2} -x^{2} + 2x - (-sin(\frac{\pi x}{2}))$ or

$Area = \int_{0}^{2} -x^{2} + 2x + sin(\frac{\pi x}{2})$

Taking the antiderivative gives

$-\frac{x^{3}}{3} + x^{2} -\frac{2}{\pi }cos(\frac{\pi x}{2})$ , evaluated from x = 0 to x = 2 .

$F(2) = \frac{-8}{3} + 4 + \frac{2 }{\pi }$ , and

$F(0) = -\frac{2}{\pi }$ .

Thus, the area is given by:

$F(2) - F(0) =- \frac{8}{3} + 4 + \frac{2}{\pi } - (-\frac{2}{\pi }) =\frac{4}{3} +\frac{4}{\pi }$

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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #12 : Determine Local Linearity And Linearization

Example Question #13 : Determine Local Linearity And Linearization

Example Question #364 : Calculus Ab

Example Question #51 : Contextual Applications Of Derivatives

Example Question #14 : Determine Local Linearity And Linearization

Example Question #1 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Example Question #2 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Example Question #3 : Analytical Applications Of Derivatives

Example Question #1 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Example Question #5 : Analytical Applications Of Derivatives

All Calculus AB Resources

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