Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #501 : Other Differential Functions

Differentiate \(\displaystyle H(x)=\int_{2}^{x} \ln (\sqrt{t^2+16}) \ dt.\) 

Possible Answers:

\(\displaystyle \ln (\sqrt{t^2+16})\)

\(\displaystyle \frac{x}{x^2+16}\)

\(\displaystyle \ln (\sqrt{x^2+16}) - \ln (\sqrt{20})\)

\(\displaystyle \ln (\sqrt{x^2+16})\)

Correct answer:

\(\displaystyle \ln (\sqrt{x^2+16})\)

Explanation:

By Fundamental Theorem of Calculus,

 \(\displaystyle \frac{dH}{dx}=\frac{\mathrm{d} }{\mathrm{d} x} \int_{2}^{x} \ln (\sqrt{t^2+16}) \ dt = H'(x) \cdot \frac{d}{dx}x = \ln(\sqrt{x^2+16}) \cdot 1 =\ln(\sqrt{x^2+16}).\)

Example Question #502 : Other Differential Functions

Find the derivative of the function

 \(\displaystyle f(x) = \sin x + x^2 + \frac{5}{27} e^2\).

Possible Answers:

\(\displaystyle - \cos x + 2x\)

\(\displaystyle \cos x + 2x\)

\(\displaystyle - \cos x +2x + \frac{10}{27}e\)

\(\displaystyle \cos x + 2x + \frac{10}{27}e\)

Correct answer:

\(\displaystyle \cos x + 2x\)

Explanation:

Since \(\displaystyle \frac{5}{27}e^2\) is a constant, its derivative is \(\displaystyle 0\). Therefore, \(\displaystyle f'(x) = \cos x + 2x.\)

Example Question #505 : How To Find Differential Functions

Find two numbers which add up to fifty such that their product is the maximum value possible.

Possible Answers:

\(\displaystyle 0,50\)

\(\displaystyle 20,30\)

\(\displaystyle 15,35\)

\(\displaystyle 25,25\)

\(\displaystyle 5,45\)

Correct answer:

\(\displaystyle 25,25\)

Explanation:

To begin with write the information of the problem in mathematical terms. We're told the sum of two numbers is fifty:

\(\displaystyle x+y=50\)

Their product is then:

\(\displaystyle f(x,y)=xy\)

We're uncertain currently about what these two numbers are, but we can relate them:

\(\displaystyle y=50-x\)

So that we can simplify the product function:

\(\displaystyle f(x)=x(50-x)\)

\(\displaystyle f(x)=50x-x^2\)

Maxima and minima occur where the derivative of a function is zero. Take the derivative of this function:

\(\displaystyle f'(x)=50-2x\)

And set it to zero:

\(\displaystyle 50-2x=0\)

\(\displaystyle -2x=-50\)

\(\displaystyle x=25\)

\(\displaystyle y=50-x=50-25=25\)

You can validate that this is a minimum due to the derivative function having positive values before this point, and negative values after this point.

Example Question #507 : How To Find Differential Functions

Let \(\displaystyle f(x)=(x^2+1)^5\) on the interval \(\displaystyle (0,2)\). Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Possible Answers:

\(\displaystyle 0.299\)

\(\displaystyle 0.753\)

\(\displaystyle 1.119\)

\(\displaystyle 1.484\)

\(\displaystyle 1.986\)

Correct answer:

\(\displaystyle 1.484\)

Explanation:

The mean value theorem states that for a planar arc passing through a starting and endpoint \(\displaystyle (a,b);a< b\), there exists at a minimum one point, \(\displaystyle c\), within the interval \(\displaystyle (a,b)\) for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

\(\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a< c< b\)

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of

\(\displaystyle f(x)=(x^2+1)^5\) on the interval \(\displaystyle (0,2)\)

\(\displaystyle f(0)=(0^2+1)^5=1\)

\(\displaystyle f(2)=(2^2+1)^5=3125\)

Then take the difference of the two and divide by the interval.

\(\displaystyle \frac{3125-1}{2-0}=1562\)

Now find the derivative of the function; this will be solved for the value(s) found above.

\(\displaystyle f'(x)=5(2x)(x^2+1)^4\)

\(\displaystyle 5(2x)(x^2+1)^4=1562\)

\(\displaystyle x=1.484\)

This value valls within the interval \(\displaystyle (0,2)\), validating the mean value theorem.

Example Question #506 : How To Find Differential Functions

Let \(\displaystyle f(x)=15x^3+23x^2-8x\) on the interval \(\displaystyle (1,1.01)\). Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Possible Answers:

\(\displaystyle -3.119\)

\(\displaystyle -2.027\)

\(\displaystyle 1.002\)

\(\displaystyle 1.008\)

\(\displaystyle 1.005\)

Correct answer:

\(\displaystyle 1.005\)

Explanation:

The mean value theorem states that for a planar arc passing through a starting and endpoint \(\displaystyle (a,b);a< b\), there exists at a minimum one point, \(\displaystyle c\), within the interval \(\displaystyle (a,b)\) for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

\(\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a< c< b\)

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of

\(\displaystyle f(x)=15x^3+23x^2-8x\) on the interval \(\displaystyle (1,1.01)\)

\(\displaystyle f(1)=15(1)^3+23(1)^2-8(1)=30\)

\(\displaystyle f(1.01)=15(1.01)^3+23(1.01)^2-8(1.01)=30.836815\)

Then take the difference of the two and divide by the interval.

\(\displaystyle \frac{30.836815-30}{1.01-1}=83.6815\)

Now find the derivative of the function; this will be solved for the value(s) found above.

\(\displaystyle f'(x)=45x^2+46x-8\)

\(\displaystyle 45x^2+46x-8=83.6815\)

\(\displaystyle x=-2.02722,1.005\)

To validate the mean value theorem, a solution must fall within the interval used, and \(\displaystyle 1.005\) does in fact fall within \(\displaystyle (1,1.01)\)

Example Question #503 : Other Differential Functions

Find the derivative. 

\(\displaystyle 2x^2+4x-6\)

Possible Answers:

\(\displaystyle 2x+4\)

\(\displaystyle 4x+x\)

\(\displaystyle 2x+x\)

\(\displaystyle 4x+4\)

Correct answer:

\(\displaystyle 4x+4\)

Explanation:

Use the power rule to find the derivative.

\(\displaystyle \frac{d}{dx}2x^2=4x\)

\(\displaystyle \frac{d}{dx}4x=4\)

\(\displaystyle \frac{d}{dx}-6=0\)

Thus, the derivative is \(\displaystyle 4x+4\)

Example Question #504 : Other Differential Functions

Find the derivative.

\(\displaystyle 6x^3\)

Possible Answers:

\(\displaystyle 6x^2\)

\(\displaystyle 18x\)

\(\displaystyle 12x\)

\(\displaystyle 18x^2\)

Correct answer:

\(\displaystyle 18x^2\)

Explanation:

Use the power rule to find the derivative. 

\(\displaystyle \frac{d}{dx}6x^3=18x^2\)

Example Question #691 : Functions

Find the derivative at x=3.

\(\displaystyle 6x^3\)

Possible Answers:

\(\displaystyle 18\)

\(\displaystyle 162\)

\(\displaystyle 204\)

\(\displaystyle 64\)

Correct answer:

\(\displaystyle 162\)

Explanation:

First, find the derivative using the power rule:

\(\displaystyle \frac{d}{dx}6x^3=18x^2\)

Now, substitute 3 for x.

\(\displaystyle 18(3^2)=162\)

Example Question #512 : How To Find Differential Functions

Let \(\displaystyle f(x)=9^{\cos(x^2)}\) on the interval \(\displaystyle (0,0.3)\). Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Possible Answers:

\(\displaystyle 0.29\)

\(\displaystyle 0.19\)

\(\displaystyle -3.10\)

\(\displaystyle 0.02\)

\(\displaystyle 1.69\)

Correct answer:

\(\displaystyle 0.19\)

Explanation:

The mean value theorem states that for a planar arc passing through a starting and endpoint \(\displaystyle (a,b);a< b\), there exists at a minimum one point, \(\displaystyle c\), within the interval \(\displaystyle (a,b)\) for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

\(\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a< c< b\)

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of

\(\displaystyle f(x)=9^{\cos(x^2)}\) on the interval \(\displaystyle (0,0.3)\)

\(\displaystyle f(0)=9^{\cos(0^2)}=9\)

\(\displaystyle f(0.3)=9^{\cos(0.3^2)}=8.9203\)

Then take the difference of the two and divide by the interval.

\(\displaystyle \frac{8.9203-9}{0.3-0}=-0.266\)

Now find the derivative of the function; this will be solved for the value(s) found above.

Derivative of an exponential: 

\(\displaystyle d[a^u]=a^udu\ln(a)\)

Trigonometric derivative: 

\(\displaystyle d[\cos(u)]=-\sin(u)du\)

\(\displaystyle f'(x)=-2x\sin(x^2)9^{\cos(x^2)}\ln(9)\)

\(\displaystyle -2x\sin(x^2)9^{\cos(x^2)}\ln(9)=-0.266\)

Using a solver, there are multiple solutions which satisfy this equation:

\(\displaystyle x=-3.54,-3.10,-2.51,-1.85,0.19,1.69,2.51,3.04\)

However, to satisfy the mean value theorem, a solution must fall within the specified interval. \(\displaystyle 0.19\) falls within \(\displaystyle (0,0.3)\).

Example Question #513 : How To Find Differential Functions

As per the mean value theorem, there exists at least one value \(\displaystyle x=c\) within the interval \(\displaystyle (2,b)\) such that \(\displaystyle f'(c)=-1\). Find \(\displaystyle b\) if \(\displaystyle f(x)=10x-x^2\) 

Possible Answers:

\(\displaystyle 4\)

\(\displaystyle 2\)

\(\displaystyle 9\)

\(\displaystyle 7\)

\(\displaystyle 5\)

Correct answer:

\(\displaystyle 9\)

Explanation:

The mean value theorem states that for a planar arc passing through a starting and endpoint \(\displaystyle (a,b);a< b\), there exists at a minimum one point, \(\displaystyle c\), within the interval \(\displaystyle (a,b)\) for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

\(\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a< c< b\)

Using our interval and function definitions, along with the derivative value of \(\displaystyle f'(c)\) given, we'll in turn wish to solve for \(\displaystyle b\) to validate the mean value theorem:

\(\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}\)

\(\displaystyle -1=\frac{(10b-b^2)-(10(2)-2^2)}{b-2}\)

\(\displaystyle 2-b=10b-b^2-16\)

\(\displaystyle b^2-11b+18=0\)

\(\displaystyle b=9\)

(Note that the solution \(\displaystyle b=2\) would not work because it would eliminate the interval as \(\displaystyle a=2\), and it would lead to a zero denominator in the working equation.)

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