the function g is defined and differentiable on the closed interval -7,5 and satisfies g(0)=5. the graph of…

Explanation:

Step1: Recall the fundamental - theorem of calculus

If (g(x)=g(0)+\int_{0}^{x}g^{\prime}(t)dt), and (g(0) = 5).

Step2: Calculate (g(3))

To find (g(3)), we use (g(3)=g(0)+\int_{0}^{3}g^{\prime}(x)dx). The integral (\int_{0}^{3}g^{\prime}(x)dx) is the area under the curve (y = g^{\prime}(x)) from (x = 0) to (x = 3). The region from (x=0) to (x = 3) is a semi - circle with radius (r = 1) and a right - triangle with base (b = 2) and height (h=2). The area of the semi - circle is (A_{1}=\frac{1}{2}\pi r^{2}=\frac{\pi}{2}), and the area of the triangle is (A_{2}=\frac{1}{2}\times2\times2 = 2). So (\int_{0}^{3}g^{\prime}(x)dx=\frac{\pi}{2}+2). Then (g(3)=5+\frac{\pi}{2}+2=7+\frac{\pi}{2}).

Step3: Calculate (g(-2))

To find (g(-2)), we use (g(-2)=g(0)+\int_{0}^{-2}g^{\prime}(x)dx=5-\int_{-2}^{0}g^{\prime}(x)dx). The region from (x=-2) to (x = 0) is a right - triangle with base (b = 2) and height (h = 1). The area of the triangle is (A=\frac{1}{2}\times2\times1 = 1). So (g(-2)=5 - 1=4).

Step4: Find points of inflection for (y = g(x))

Points of inflection occur where (g^{\prime\prime}(x)=0) or (g^{\prime\prime}(x)) does not exist. Since (g^{\prime\prime}(x)) is the derivative of (g^{\prime}(x)), we look for where the slope of (y = g^{\prime}(x)) changes sign. The slope of (g^{\prime}(x)) changes sign at (x=-1,1,3). So the (x) - coordinates of the points of inflection of (y = g(x)) are (x=-1,1,3).

Step5: Find critical points of (h(x))

First, find the derivative of (h(x)). Since (h(x)=g(x)-\frac{1}{2}x^{2}), then (h^{\prime}(x)=g^{\prime}(x)-x). Critical points of (h(x)) occur where (h^{\prime}(x)=0), i.e., (g^{\prime}(x)=x). We find the intersection points of (y = g^{\prime}(x)) and (y = x) on the interval (-7\lt x\lt5). For (x\in[-7, - 1]), (g^{\prime}(x)=\frac{1}{6}(x + 7)-1=\frac{x+7 - 6}{6}=\frac{x + 1}{6}). Setting (\frac{x + 1}{6}=x), we get (x+1 = 6x), (5x=1), (x=\frac{1}{5}\notin[-7,-1]). For (x\in[-1,1]), (g^{\prime}(x)) is the upper - half of a circle (y=\sqrt{1 - x^{2}}). Setting (\sqrt{1 - x^{2}}=x), we square both sides (1 - x^{2}=x^{2}), (2x^{2}=1), (x=\pm\frac{\sqrt{2}}{2}). Since (x\in[-1,1]), (x=\frac{\sqrt{2}}{2}) is a solution. For (x\in[1,3]), (g^{\prime}(x)=x - 1). Setting (x - 1=x), no solution. For (x\in[3,5]), (g^{\prime}(x)=-\frac{3}{2}(x - 3)+3=-\frac{3}{2}x+\frac{9 + 6}{2}=-\frac{3}{2}x+\frac{15}{2}). Setting (-\frac{3}{2}x+\frac{15}{2}=x), we get (-3x + 15 = 2x), (5x=15), (x = 3). To classify the critical points, we use the second - derivative test. (h^{\prime\prime}(x)=g^{\prime\prime}(x)-1). At (x=\frac{\sqrt{2}}{2}), (g^{\prime\prime}(x)) is positive (the slope of (g^{\prime}(x)) is increasing), so (h^{\prime\prime}(\frac{\sqrt{2}}{2})=g^{\prime\prime}(\frac{\sqrt{2}}{2})-1). Since (g^{\prime\prime}(\frac{\sqrt{2}}{2})\gt0), (h^{\prime\prime}(\frac{\sqrt{2}}{2})\gt - 1). Also, we can use the first - derivative test. For (x) slightly less than (\frac{\sqrt{2}}{2}), (g^{\prime}(x)\gt x) (by looking at the graph of (y = g^{\prime}(x)) and (y = x)), (h^{\prime}(x)\gt0), and for (x) slightly greater than (\frac{\sqrt{2}}{2}), (g^{\prime}(x)\lt x), (h^{\prime}(x)\lt0). So (x=\frac{\sqrt{2}}{2}) is a relative maximum. At (x = 3), (g^{\prime\prime}(x)) changes from positive to negative (the slope of (g^{\prime}(x)) changes from positive to negative), (h^{\prime\prime}(3)=g^{\prime\prime}(3)-1\lt0), so (x = 3) is a relative maximum.

Answer:

(a) (g(3)=7+\frac{\pi}{2}), (g(-2)=4) (b) (x=-1,1,3) (c) Critical points are (x=\frac{\sqrt{2}}{2},3). Both are relative maxima.