Advanced Placement (AP), 15.08.2021 07:10 ricardodeleon152

CALCULUS AB Given g(x), name the piecewise function and graph:
g(x)+1
g(x-1)
-g(x)
2g(x)
-g(x-1)
|g(x)|

CALCULUS AB

Given g(x), name the piecewise function and graph:g(x)+1g(x-1)-g(x)2g(x)-g(x-1)|g(x)|

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Answer from: budjasdatazaki3913

A piecewise function is one that has different operations, depending on the values of the input

The piecewise function is;

$g(x) = \begin{cases} -\left | x + 2 \right |&\mathbf{ \text{ if } -5 \leq x < 0 }\\ 2 \cdot x - 2 &\mathbf{ \text{ if } 0 < x \leq 2}\\ 2 &\mathbf{ \text{ if } 2 \leq x \leq 5}\end{cases}$

Reason:

The general form of the absolute value function is presented as follows;

$f(x) = \mathbf{a \left | x - h \right | + k}$

The function given in the graph are;

Domain; -5 ≤ x < 0

Function; $g(x) = -\left | x + 2 \right |$

Domain; 0 < x ≤ 2

Points on the graph are;

(0, -2), and (2, 2)

Slope = (2 - (-2))/(2 - 0) = 2

g(x) - (-2) = 2·(x - 0)

∴ Function; g(x) = 2·x - 2

Domain; 2 ≤ x ≤ 5

Function; g(x) = 2

Therefore, we have the following piecewise function;

$g(x) = \mathbf{ \begin{cases} -\left | x + 2 \right |& \text{ if } -5 \leq x < 0 \\ 2 \cdot x - 2 & \text{ if } 0 < x \leq 2\\ 2 & \text{ if } 2 \leq x \leq 5\end{cases}}$

Learn more about piecewise functions here:

Answer from: Quest

it is snorkeling, apex verified.

Answer from: Quest

Answer: the interval on which the curve is concave up is (-∞, -1/2)====== $\displaystyle y = \int_{0}^x \frac{1}{1 + t + t^2} \; dt$ we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \end{aligned}$ let the function f denote be the anti-derivative of the inside expression, i.e. a function such that $(d/dx)f(x) = 1/(1 + t + t^2)$ .then use the fundamental theorem of calculus: $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \\& = \frac{d}{dx} \big[ f(x) - f(0) \big] & & \text{\footnotesize apply ftc} \\ & = \frac{d}{dx}f(x) - 0 \\ & = \frac{1}{x + x + x^2} & & \text{\footnotesize since $\frac{d}{dx}f(x) = \frac{1}{1 + x + x^2}$ } \end{aligned}$ differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function $\begin{aligned} \frac{d}{dx} \left( \frac{dy}{dx} \right) & = \frac{d}{dx}\left(\frac{1}{1 + x + x^2} \right) \\ \frac{d^2y}{dx^2} & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot \tfrac{d}{dx}(1 + x + x^2) \\ & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot (1 + 2x) \\ & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \end{aligned}$ find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x $\begin{aligned} 0 & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \\ 0 & = -(1+2x) \\ -1 & = 2x \\ x & = -1/2 \end{aligned}$ where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation 1 + x + x² = 0results in a discriminant of $b^2 - 4ac = 1^2 - 4(1)(1) = -3$ meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = -1} & = \frac{-(1+2(-1))}{\left(1+(-1)+(-1)^2\right)^2} \\ & = \frac{-(1-2)}{1} \\ & = 1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = 0} & = \frac{-(1+2(0))}{\left(1+(0)+(0)^2\right)^2} \\ & = \frac{-(1)}{1} \\ & = -1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is (-∞, -1/2)

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CALCULUS AB Given g(x), name the piecewise function and graph:
g(x)+1
g(x-1)
-g(x)