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3 years ago

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Find the absolute extrema if they exist, as well as all values of x where they occur, for the function f (x )equals2 x Supersc

ript 4 Baseline minus 36 x squared plus 2 on the domain [negative 4 comma 4 ].
Find the derivative of f left parenthesis x right parenthesisequals2 x cubed plus 20 x squared plus 50 x plus 1. f prime left parenthesis x right parenthesisequals nothing Identify the absolute maximum if it exists, as well as all values of x where it occurs.

1 answer:

kkurt [141]3 years ago

5 0

Step-by-step explanation:

Derivative of the first function:

$f(x)=2x^{4} -36x^{2} +2 on D:[-4,4]$

$f'(x)=8x^{3} -72x$

set that equal to 0 and solve for possible x values.

$8x(x^{2} -9)=0\\x=0, -3, 3$

put x back in to original equation to get a y value

y = 2, -160, -160,

Absolute min at (-3, -160) and (3, 160) and an absolute max at (0,2)

Second Part

Find derivative of $f(x)=2x^{3} +20x^{2} +50x +1$

$f'(x)=6x^{2} +40x+50$

set equal to zero, it is a quadratic so you can use the quadratic formula to solve for x

x= -5 or -5/3

put x back into the original function to get a y value

y = 1 or -973/27

so absolute max at (-5, 1) and absolute min at (-5/3, -973/27)

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3 years ago

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

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Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

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$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

$2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}$

Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

If $n=2$ , then

$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

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