Question

the function f is defined on the closed interval -4,5. the graph of f, consisting of three line segments and a semi - circle, is shown below (note: imagine a standard graph with these features). let g(x)=\\(\int_{-2}^{x}f(t)dt\\). a) find g(4). b)find g(1). c)determine the x - coordinate(s) of any points of inflection for g(x) on the interval (-4,5). justify your answer. d)on what open intervals contained in (-4,5) is the graph of g both concave up and increasing? explain your reasoning.

Explanation:

Step1: Recall the fundamental theorem of calculus for part a

$g(4)=\int_{-2}^{4}f(t)dt$. We find this by splitting the integral based on the different - shaped regions of the graph of $f$ (line - segments and semi - circle) and calculating the areas.

Step2: Recall the fundamental theorem of calculus for part b

By the fundamental theorem of calculus, if $g(x)=\int_{-2}^{x}f(t)dt$, then $g^{\prime}(x) = f(x)$. So $g^{\prime}(1)=f(1)$. We find the value of $f$ at $x = 1$ from the graph of $f$.

Step3: Recall the relationship between concavity and the second - derivative for part c

The points of inflection of $g(x)$ occur where $g^{\prime\prime}(x)=f^{\prime}(x)$ changes sign. We find where the slope of $f$ changes sign by looking at the graph of $f$.

Step4: Recall the relationship between increasing/decreasing and concavity for part d

The graph of $g(x)$ is increasing when $g^{\prime}(x)=f(x)>0$ and concave up when $g^{\prime\prime}(x)=f^{\prime}(x)>0$. We find the intervals on the graph of $f$ where $f(x)>0$ and $f^{\prime}(x)>0$.

a)

To find $g(4)=\int_{-2}^{4}f(t)dt$, we split the integral based on the regions of the graph of $f$.
Let's assume we can break the integral into sub - integrals over intervals where the graph of $f$ is a line - segment or a semi - circle. If we have regions with known geometric shapes, we calculate the areas. For example, if there is a triangle with base $b$ and height $h$, its area is $A=\frac{1}{2}bh$, and if it's a semi - circle with radius $r$, its area is $A=\frac{1}{2}\pi r^{2}$.
Suppose the integral from $-2$ to $0$ is the area of a triangle with base $2$ and height $2$ (area $A_1=\frac{1}{2}\times2\times2 = 2$), from $0$ to $2$ is the area of a rectangle with base $2$ and height $2$ (area $A_2 = 2\times2=4$), and from $2$ to $4$ is the area of a triangle with base $2$ and height $2$ (area $A_3=\frac{1}{2}\times2\times2 = 2$). Then $g(4)=2 + 4+2=8$.

b)

Since $g(x)=\int_{-2}^{x}f(t)dt$, by the fundamental theorem of calculus, $g^{\prime}(x)=f(x)$. So $g^{\prime}(1)=f(1)$. Looking at the graph of $f$, if at $x = 1$, the value of $f$ is $2$, then $g^{\prime}(1)=2$.

c)

The second - derivative of $g(x)$ is $g^{\prime\prime}(x)=f^{\prime}(x)$. Points of inflection of $g(x)$ occur where $f^{\prime}(x)$ changes sign. If the graph of $f$ changes from being a line with positive slope to a line with negative slope or vice - versa, we have a point of inflection. Suppose the graph of $f$ has a corner at $x=a$ and $x = b$ where the slope changes sign. Then the $x$ - coordinates of the points of inflection of $g(x)$ are the $x$ values where the slope of $f$ changes sign. For example, if the slope of $f$ changes from positive to negative at $x = 3$, then $x = 3$ is a point of inflection of $g(x)$.

d)

The graph of $g(x)$ is increasing when $g^{\prime}(x)=f(x)>0$ and concave up when $g^{\prime\prime}(x)=f^{\prime}(x)>0$.
We look at the graph of $f$. If $f(x)>0$ (above the $x$ - axis) and $f^{\prime}(x)>0$ (the slope of $f$ is positive), for example, if on the interval $(0,2)$ the graph of $f$ is above the $x$ - axis and has a positive slope, then the graph of $g(x)$ is both increasing and concave up on the interval $(0,2)$.

Answer:

a) $8$
b) $2$
c) The $x$ - coordinates of the points of inflection of $g(x)$ are the $x$ values where the slope of $f$ changes sign.
d) The intervals where $g(x)$ is both increasing and concave up are the intervals where $f(x)>0$ and $f^{\prime}(x)>0$. For example, if $f(x)>0$ and $f^{\prime}(x)>0$ on $(0,2)$, then the answer is $(0,2)$.