Introductory Calculus

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Find the derivative of the following function

$f'(x)=-12x^4+18x^3-2x^2+8x\textup{}$

$f'(x)=-\frac{3}{4}x^3+2x^2-\frac{1}{2}x+8$

Explanation

To find the derivative of this function, we simply need to use the Power Rule. The Power rule states that for each term, we simply multiply the coefficient by the power to find the new coefficient. We then decrease the power by one to obtain the degree of the new term.

For example, with our first term, , we would multiply the coefficient by the power to obtain the new coefficient of . We then decrease the power by one from 4 to 3 for the new degree. Therefore, our new term is . We then simply repeat the process with the remaining terms.

Note that with the second to last term, our degree is 1. Therefore, multiplying the coefficient by the power gives us the same coefficient of 8. When the degree decreases by one, we have a degree of 0, which simply becomes 1, making the entire term simply 8.

With our final term, we technically have

Therefore, multiplying our coefficient by our power of 0 makes the whole term 0 and thus negligible.

Our final derivative then is

Determine the values for the points of inflection of the following function:

$\frac{1}{3}$

$\frac{-1}{3}$

Explanation

To solve, you must set the second derivative equal to 0 and solve for x. To differentiate twice, use the power rule as outlined below:

Power Rule: $(x^n)'=nx^{n-1}$

Therefore:

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

Remember, the derivative a constant is 0.

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

Now, set it equal to 0. Thus,

$x=\frac{2}{6}$

$x=\frac{1}{3}$

Find the derivative of the following function

$f'(x)=-12x^4+18x^3-2x^2+8x\textup{}$

$f'(x)=-\frac{3}{4}x^3+2x^2-\frac{1}{2}x+8$

Explanation

With our final term, we technically have

Therefore, multiplying our coefficient by our power of 0 makes the whole term 0 and thus negligible.

Our final derivative then is

Find where .

Explanation

In order to find the derivative we will need to use the power rule on each term. The power rule states,

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

Applying this rule we get the following.

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

Given the following function, find the critical numbers:

$\frac{1}{3}$

$-\frac{1}{3}$

$-\frac{2}{3}$

Explanation

Critical numbers are where the slope of the function is equal to zero or undefined.

Find the derivative and set the derivative function to zero.

$x=\frac{1}{3}$

There is only one critical value at $x=\frac{1}{3}$ .

Determine the x-coordinate of the inflection point of the function .

$\frac{4}{3}$

$-\frac{4}{3}$

$\frac{8}{3}$

Explanation

The point of inflection exists where the second derivative is zero.

, and we set this equal to zero.

$x=\frac{8}{6}=\frac{4}{3}$

Find the first derivative of $y=x^{3}$ .

$\frac{dy}{dx}=3x^{2}$

$\frac{dy}{dx}=3x$

$\frac{dx}{dy}=3x$

$y=3x^{2}$

Explanation

By the Power Rule of derivatives, for any equation $y=x^{a}$ , the derivative $\frac{dy}{dx}=a\cdot x^{a-1}$ .

With our function $y=x^{3}$ , where , we can therefore conclude that:

$\frac{dy}{dx}=3\cdot x^{3-1}$

$\frac{dy}{dx}=3x^{2}$

Find the derivative of the function $f(x)=x^{3}+5$ .

$f'(x)=3x^{2}$

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

$f'(x)=3x^{3}$

$f'(x)=\frac{1}4{}x^{4}$

None of the above

Explanation

For any function $f(x)=Ax^{n}$ , the first derivative $f'(x)=nAx^{n-1}$ .

Therefore, taking each term of :

$f(x)=x^{3}+5$

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

$f'(x)=3x^{2}$

Find where .

Explanation

In order to find the derivative we will need to use the power rule on each term. The power rule states,

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

Applying this rule we get the following.

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

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

$\sin 1$

$\infty$

The limit does not exist.

Explanation

This can be rewritten as follows:

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

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

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

$= \lim_{x\rightarrow \infty } \frac{ \sin \frac{1}{ x^{2}-1} }{\frac{1}{ x^{2}-1}}$

We can substitute $u = \frac{1}{ x^{2}-1}$ , noting that as $x\rightarrow \infty$ , $u \rightarrow 0$ :

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

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