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3 years ago

Complete the square to determine the minimum or maximum value of the function defined by the expression.

Mathematics

1 answer:

Alekssandra [29.7K]3 years ago

3 0

The minimum value at -1 ⇒ D

Step-by-step explanation:

The completing square form of ax² + bx + c is a(x - h)² + k, where

$h=\frac{-b}{2a}$
k is the value of of the expression when x = h
k is minimum if a > 0 and maximum if a < 0

∵ The expression is x² + 4x + 3

∴ a = 1 , b = 4 , c = 3

- Use the rule above to find h

∵ $h=\frac{-4}{2(1)}$

∴ h = -2

- To find k substitute x by the value of h

∵ k = (-2)² + 4(-2) + 3 = 4 - 8 + 3

∴ k = -1

- Substitute h and k in the form of the completing square

∵ a(x - h)² + k = 1(x - -2)² + (-1)

∴ a(x - h)² + k = (x + 2)² - 1

∴ x² + 4x + 3 = (x + 2)² - 1

∵ The completing square is (x + 2)² - 1

∵ a = 1 ⇒ greater than zero

∴ The value is minimum

- The minimum value is the value of k

∵ k = -1

∴ The minimum value of the function is -1

The minimum value at -1

Learn more:

You can learn more about the quadratic function in brainly.com/question/9390381

#LearnwithBrainly

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Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

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The left and right endpoints of the $i$ -th subinterval, respectively, are

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$r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8$

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We approximate the (signed) area under the curve over each subinterval by

$T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

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We approximate the area for each subinterval by

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We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

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