answer:To differentiate {eq}g(x) = x sqrt{2x^2-1} {/eq}, we'll apply the product rule. [ frac{dg}{dx} = frac{d}{dx}(x sqrt{2x^2-1}) ] Using the product rule: [ frac{dg}{dx} = sqrt{2x^2 - 1} frac{d}{dx}(x) + x frac{d}{dx}(sqrt{2x^2-1}) ] Now apply the derivative of x and the chain rule for the second term: [ frac{dg}{dx} = sqrt{2x^2 - 1} + x cdot frac{1}{2sqrt{2x^2-1}} cdot frac{d}{dx}(2x^2-1) ] The derivative of {eq}2x^2-1 {/eq} is {eq}4x {/eq}: [ frac{dg}{dx} = sqrt{2x^2 - 1} + frac{2x^2}{sqrt{2x^2-1}} ] Combine the terms: [ frac{dg}{dx} = frac{4x^2 - 1}{sqrt{2x^2-1}} ] Therefore, the derivative of {eq}g(x) {/eq} is: [ g'(x) = boxed{frac{4x^2 - 1}{sqrt{2x^2-1}}} ]

answer:GDP per capita is a measure of the total output of a country divided by its population, providing an indication of the average income earned by individuals. When GDP per capita is higher, it suggests that people in that economy are producing more and earning higher incomes, which generally corresponds to a higher standard of living. Conversely, a lower GDP per capita implies that people are producing less and earning less, leading to a lower standard of living. Poverty rates provide another perspective on the standard of living. A higher poverty rate indicates that a larger proportion of the population is living below the poverty line, struggling to meet basic needs. Conversely, a lower poverty rate implies that more people have access to resources necessary for a decent standard of living. In summary, a country with a higher GDP per capita and a lower poverty rate is likely to have a higher standard of living, reflecting better economic conditions and opportunities for its citizens.

answer:1. First, we can simplify the square root terms: 4√2 = 4 * 1.414 = 5.656 13/√2 = 13 / 1.414 = 9.200 2. Now we have the following numbers: {-6, -4, -4, -3.76, 2, 5, 5.656, 9.200} 3. We can sort these numbers in ascending order: {-6, -4, -4, -3.76, 2, 5, 5.656, 9.200} The answer is {-6, -4, -4, -3.76, 2, 5, 4√2, 13/√2}

answer:To find the points of concavity change, we need to find the second derivative of the function and equate it to zero: begin{align*} f(x) &= x^4 - 8x^2 f'(x) &= 4x^3 - 16x f''(x) &= 12x^2 - 16 end{align*} Setting {eq}f''(x) = 0{/eq}, we get: begin{align*} 12x^2 - 16 &= 0 x^2 &= frac{4}{3} x &= pm frac{2}{sqrt{3}} end{align*} These are the points of concavity change. To determine the intervals of concavity, we can use a sign chart for {eq}f''(x){/eq}: | Interval | {eq}f''(x){/eq} | Concavity | |---|---|---| | {eq}(-infty, -frac{2}{sqrt{3}}){/eq} | + | Concave up | | {eq}(-frac{2}{sqrt{3}}, frac{2}{sqrt{3}}){/eq} | - | Concave down | | {eq}(frac{2}{sqrt{3}}, infty){/eq} | + | Concave up | Therefore, the function is concave up on the intervals {eq}(-infty, -frac{2}{sqrt{3}}){/eq} and {eq}(frac{2}{sqrt{3}}, infty){/eq}, and concave down on the interval {eq}(-frac{2}{sqrt{3}}, frac{2}{sqrt{3}}){/eq}.