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High School Math : High School Math

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Example Questions

Example Question #3 : Calculus I — Derivatives

Calculate $\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$ .

Possible Answers:

$\frac{1}{2}$

The limit does not exist.

Correct answer:

Explanation:

Substitute $u = 2x^{2} - 2$ to rewrite this limit in terms of u instead of x. Multiply the top and bottom of the fraction by 2 in order to make this substitution:

(Note that as $x \rightarrow1$ , $u \rightarrow 0$ .)

$\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$

$=2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2\left (x^{2}-1 \right )}$

$= 2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2x^{2}-2 \right )}$

$= 2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u}$

$\lim_{u \rightarrow 0} \frac{\sin u }{u} = 1$ , so

$2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u} = 2\cdot1 = 2$ , which is therefore the correct answer choice.

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Example Question #4 : Calculus I — Derivatives

Calculate $\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

Possible Answers:

The limit does not exist.

Correct answer:

Explanation:

You can substitute $u = 4x^{2}$ to write this as:

$\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

$= \lim_{x \rightarrow 0}\left (4 \cdot \frac{\sin 4 x^{2} }{4x^{2}} \right )$

$=4 \lim_{x \rightarrow 0} \frac{\sin 4 x^{2} }{4x^{2}}$

$=4 \lim_{u \rightarrow 0} \frac{\sin u }{u}$

Note that as $x \rightarrow 0$ , $u = 4x^{2} \rightarrow 0$

$\lim_{u \rightarrow 0} \frac{\sin u }{u}$ , since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

$\lim_{u \rightarrow 0} \frac{\sin u }{u}=\frac{(\sin u)'}{u'}=\frac{\cos u}{1}=\cos u$

$4\lim_{u\rightarrow 0}\cos u=4\cdot \cos(0)=4\cdot 1$

$4 \lim_{u \rightarrow 0} \frac{\sin u }{u} = 4 \cdot 1 = 4$ , which is the correct choice.

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Example Question #1 : Derivatives

The speed of a car traveling on the highway is given by the following function of time:

v(t) = 5t^2+2t

Note that

v(0) =0

What does this mean?

Possible Answers:

The car's speed is constantly changing at time t=0 .

The car takes seconds to reach its maximum speed.

The car is not accelerating at time t=0 .

The car is not moving at time t=0 .

The car is not decelerating at time t=0 .

Correct answer:

The car is not moving at time t=0 .

Explanation:

The function v(t) gives you the car's speed at time . Therefore, the fact that v(0)=0 means that the car's speed is at time . This is equivalent to saying that the car is not moving at time t=0 . We have to take the derivative of to make claims about the acceleration.

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Example Question #2 : Derivatives

The speed of a car traveling on the highway is given by the following function of time:

f(t) = 5t^2+2t

Consider a second function:

g(t) = 10t+2

What can we conclude about this second function?

Possible Answers:

It represents the change in distance over a given time .

It represents another way to write the car's speed.

It represents the total distance the car has traveled at time .

It has no relation to the previous function.

It represents the rate at which the speed of the car is changing.

Correct answer:

It represents the rate at which the speed of the car is changing.

Explanation:

Notice that the function g(t) is simply the derivative of f(t) with respect to time. To see this, simply use the power rule on each of the two terms.

f'(t) = (2)(5)t + 2 = g(t)

Therefore, g(t) is the rate at which the car's speed changes, a quantity called acceleration.

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Example Question #2081 : High School Math

Define $g(x) = x^{3} -9x^{2}+24x+17$ .

Give the interval(s) on which is decreasing.

Possible Answers:

$(-\infty ,2) \cup (4,\infty)$

$(-\infty ,\infty)$

$(3,\infty)$

(2,4)

$(-\infty ,3)$

Correct answer:

(2,4)

Explanation:

is decreasing on those intervals at which g'(x)<0 .

$g(x) = x^{3} -9x^{2}+24x+17$

$g'(x) = 3\cdot x^{3-1} -2\cdot 9x^{2-1}+24+0$

$g'(x) = 3x^{2} -18x+24$

We need to find the values of for which $g'(x) = 3x^{2} -18x+24 <0$ . To that end, we first solve the equation:

$3x^{2} -18x+24 =0$

$3 \left (x^{2} -6x+8 \right ) =0$

$3 \left (x-2 \right )\left (x-4 \right ) =0$

$x=2\textrm{ or } x = 4$

These are the boundary points, so the intervals we need to check are:

$(-\infty ,2)$ , (2,4) , and $(4,\infty)$

We check each interval by substituting an arbitrary value from each for .

$(-\infty ,2)$

Choose x = 0

$3x^{2} -18x+24 =3\cdot 0^{2} -18\cdot 0+24 =24 > 0$

increases on this interval.

(2,4)

Choose x = 3

$3x^{2} -18x+24 =3\cdot 3^{2} -18\cdot 3+24 =27 -54 +24 = -3 < 0$

decreases on this interval.

$(4,\infty)$

Choose x = 5

$3x^{2} -18x+24 =3\cdot 5^{2} -18\cdot 5+24 = 75 -90 +24 = 9 > 0$

increases on this interval.

The answer is that decreases on (2,4) .

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Example Question #13 : Calculus I — Derivatives

Define $f \left ( x\right ) = x^{3} - \frac{21}{2} x^{2} +30x -6$ .

Give the interval(s) on which is increasing.

Possible Answers:

$(-\infty ,\infty)$

$(-\infty ,2) \cup (5,\infty)$

(2,5)

$\left ( -\infty , 3 \frac{1}{2}\right )$

$\left ( 3 \frac{1}{2}, \infty \right )$

Correct answer:

$(-\infty ,2) \cup (5,\infty)$

Explanation:

is increasing on those intervals at which f'(x)>0 .

$f \left ( x\right ) = x^{3} - \frac{21}{2} x^{2} +30x -6$

$f' \left ( x\right ) = 3x^{3-1} - 2 \cdot \frac{21}{2} x^{2-1} +30 -0$

$f' \left ( x\right ) = 3x^{2} - 21 x +30$

We need to find the values of for which $f' \left ( x\right ) = 3x^{2} - 21 x +30 > 0$ . To that end, we first solve the equation:

$3x^{2} - 21 x +30 = 0$

$3(x^{2} - 7 x +10 )= 0$

3(x-2)(x-5)= 0

$x = 2 \textrm{ or }x = 5$

These are the boundary points, so the intervals we need to check are:

$(-\infty ,2)$ , (2,5) , and $(5,\infty)$

We check each interval by substituting an arbitrary value from each for .

$(-\infty ,2)$

Choose x = 0

$3x^{2} - 21 x +30 = 3 \cdot 0^{2} - 21 \cdot 0 +30 = 30 > 0$

increases on this interval.

(2,5)

Choose x = 3

$3x^{2} - 21 x +30 = 3 \cdot 3^{2} - 21 \cdot 3 +30 = 27 -63 +30 = -6 < 0$

decreases on this interval.

$(5,\infty)$

Choose x = 6

$3x^{2} - 21 x +30 = 3 \cdot 6^{2} - 21 \cdot 6 +30 = 108 -126 + 30=12 > 0$

increases on this interval.

The answer is that increases on $(-\infty ,2) \cup (5,\infty)$

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Example Question #2082 : High School Math

At what point does -8x^2+15 shift from increasing to decreasing?

Possible Answers:

x=-16

x=0

x=-8

$x=-\frac{1}{16}$

It does not shift from increasing to decreasing

Correct answer:

x=0

Explanation:

To find out where the graph shifts from increasing to decreasing, we need to look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as 15x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{2-1})+(0*15x^{0-1})$

Notice that $(0*15x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$

If we were to graph y=-16x , would the y-value change from positive to negative? Yes. Plug in zero for y and solve for x.

y=-16x

0=-16x

$\frac{0}{-16}=x$

0=x

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Example Question #2083 : High School Math

At what point does $4x^2+\frac{2}{3}x$ shift from decreasing to increasing?

Possible Answers:

$-\frac{1}{12}=x$

-12=x

0=x

$\frac{1}{8}=x$

$-\frac{1}{24}=x$

Correct answer:

$-\frac{1}{12}=x$

Explanation:

To find out where it shifts from decreasing to increasing, we need to look at the first derivative. The shift will happen where the first derivative goes from a negative value to a positive value.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{2-1})+(1*\frac{2}{3}x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{1})+(1*\frac{2}{3}x^{0})$

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(8x)+(1*\frac{2}{3}(1))$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=8x+\frac{2}{3}$

Can this equation be negative? Yes. Does it shift from negative to positive? Yes. Therefore, it will shift from negative to positive at the point that $0=8x+\frac{2}{3}$ .

$0=8x+\frac{2}{3}$

$-\frac{2}{3}=8x$

$\frac{-\frac{2}{3}}{8}=x$

$-\frac{1}{12}=x$

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Example Question #2084 : High School Math

At what value of does 5x^2+12x shift from decreasing to increasing?

Possible Answers:

$x=-\frac{6}{5}$

$x=\frac{5}{6}$

x=-12

x=-10

It does not shift from decreasing to increasing

Correct answer:

$x=-\frac{6}{5}$

Explanation:

To find out when the function shifts from decreasing to increasing, we look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{2-1})+(1*12x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{1})+(1*12x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12$

From here, we want to know if there is a point at which graph changes from negative to positive. Plug in zero for y and solve for x.

y=10x+12

0=10x+12

-12=10x

$\frac{-12}{10}=x$

$\frac{-6}{5}=x$

This is the point where the graph shifts from decreasing to increasing.

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Example Question #2085 : High School Math

At the point x=2 , is the function -8x^2+15 increasing or decreasing, concave or convex?

Possible Answers:

The function is undefined at that point

Decreasing, convex

Decreasing, concave

Increasing, convex

Increasing, concave

Correct answer:

Decreasing, convex

Explanation:

First, let's find out if the graph is increasing or decreasing. For that, we need the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as 15x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{2-1})+(0*15x^{0-1})$

Notice that $(0*15x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$

Plug in our given point for . If the result is positive, the function is increasing. If the result is negative, the function is decreasing.

y=-16x

y=-16(2)

y=-32

Our result is negative, therefore the function is decreasing.

To find the concavity, look at the second derivative. If the function is positive at our given point, it is concave. If the function is negative, it is convex.

To find the second derivative we repeat the process, but using -16x as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=1*-16x^{1-1}$

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16x^{0}$

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16$

As you can see, our second derivative is a constant. It doesn't matter what point we plug in for ; our output will always be negative. Therefore our graph will always be convex.

Combine our two pieces of information to see that at the given point, the graph is decreasing and convex.

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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #3 : Calculus I — Derivatives

Example Question #4 : Calculus I — Derivatives

Example Question #1 : Derivatives

Example Question #2 : Derivatives

Example Question #2081 : High School Math

Example Question #13 : Calculus I — Derivatives

Example Question #2082 : High School Math

Example Question #2083 : High School Math

Example Question #2084 : High School Math

Example Question #2085 : High School Math

All High School Math Resources

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