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Find the critical numbers of the function. (Enter your answers as a comma-separated list. Use n to denote any arbitrary integer

lyudmila [28]

If

$f(\theta)=10\cos\theta+5\sin^2\theta$

then the derivative is

$f'(\theta)=-10\sin\theta+10\sin\theta\cos\theta$

Critical points occur where $f'(\theta)=0$ . This happens for

$-10\sin\theta+10\sin\theta\cos\theta=0$

$-10\sin\theta(1-\cos\theta)=0$

$\implies-10\sin\theta=0\text{ or }1-\cos\theta=0$

In the first case, we find

$-10\sin\theta=0\implies\sin\theta=0\implies\theta=n\pi$

In the second,

$1-\cos\theta=0\implies\cos\theta=1\implies\theta=2n\pi$

So all the critical points occur at multiples of $\pi$ , or $n\pi$ . (This includes all the even multiples of $\pi$ .)

3 0

3 years ago

Write the equation of the line that passes through the points (0,2) and (-1,-1):

slava [35]

Answer:

m = (2 - (-1)) / (0 - (-1)) = 3/1 = 3

y = 3 * (x - 0) + 2

y = 3 * x + 2

6 0

3 years ago

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .