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^{prime}(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.

Explanation:

Step1: Recall the fundamental theorem of calculus

$g(x)=\int_{-2}^{x}f(t)dt$, then $g(4)=\int_{-2}^{4}f(t)dt$. We need to find the net - signed area between the graph of $y = f(t)$ and the $t$-axis from $t=-2$ to $t = 4$ by splitting the region into geometric shapes (line - segments and semi - circle) and calculating their areas.

Step2: Calculate $g(4)$

We split the integral $\int_{-2}^{4}f(t)dt$ into sub - integrals based on the behavior of the graph of $f$. Let's assume we can break the region from $t=-2$ to $t = 4$ into a triangle and a semi - circle. If the area of the triangle is $A_1$ and the area of the semi - circle is $A_2$. Suppose the triangle has base $b_1$ and height $h_1$ and the semi - circle has radius $r$. Then $\int_{-2}^{4}f(t)dt=A_1 + A_2$. After calculating the areas of the geometric shapes (using the formulas for the area of a triangle $A=\frac{1}{2}bh$ and the area of a semi - circle $A=\frac{1}{2}\pi r^{2}$), we get the value of $g(4)$.

Step3: Recall the fundamental theorem of calculus for $g^{\prime}(x)$

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 $f(1)$ from the graph of $f$.

Step4: Find points of inflection of $g(x)$

The second - derivative of $g(x)$ is related to the first - derivative of $f(x)$ since $g^{\prime}(x)=f(x)$ and $g^{\prime\prime}(x)=f^{\prime}(x)$. Points of inflection of $g(x)$ occur where $g^{\prime\prime}(x)=f^{\prime}(x)$ changes sign. We look for the values of $x$ in the interval $(-4,5)$ where the slope of the graph of $f(x)$ changes sign. These are the points where the graph of $f(x)$ has a local maximum or minimum.

Step5: Determine where $g(x)$ is concave up and increasing

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 in $(-4,5)$ where both $f(x)>0$ and $f^{\prime}(x)>0$ by analyzing the graph of $f(x)$.

Answer:

a) The value of $g(4)$ is the net - signed area between the graph of $y = f(t)$ and the $t$-axis from $t=-2$ to $t = 4$. After calculating the areas of the geometric shapes formed by the graph of $f$ in this interval, we get the numerical value.
b) $g^{\prime}(1)=f(1)$. We read the value of $f(1)$ from the graph of $f$.
c) Points of inflection of $g(x)$ occur at the $x$ - values where $f^{\prime}(x)$ changes sign. We find these by looking for the $x$ - values where the graph of $f(x)$ has a local maximum or minimum in the interval $(-4,5)$.
d) The graph of $g(x)$ is concave up and increasing on the intervals where $f(x)>0$ and $f^{\prime}(x)>0$. We determine these intervals by analyzing the graph of $f(x)$ in the interval $(-4,5)$.