let f(x)=x³ - 16x² + 39x - 47. find an open interval where f decreases. as your answer, please input a + b…

Explanation:

Step1: Find the derivative of (f(x))

Using the power - rule ((x^n)^\prime=nx^{n - 1}), if (f(x)=x^{3}-16x^{2}+39x - 47), then (f^\prime(x)=3x^{2}-32x + 39).

Step2: Find the critical points

Set (f^\prime(x)=0), so (3x^{2}-32x + 39 = 0). Using the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for (ax^{2}+bx + c = 0), here (a = 3), (b=-32), (c = 39). Then (x=\frac{32\pm\sqrt{(-32)^{2}-4\times3\times39}}{2\times3}=\frac{32\pm\sqrt{1024 - 468}}{6}=\frac{32\pm\sqrt{556}}{6}=\frac{32\pm2\sqrt{139}}{6}=\frac{16\pm\sqrt{139}}{3}). (x_1=\frac{16+\sqrt{139}}{3}\approx\frac{16 + 11.79}{3}=\frac{27.79}{3}\approx9.26) and (x_2=\frac{16-\sqrt{139}}{3}\approx\frac{16 - 11.79}{3}=\frac{4.21}{3}\approx1.40).

Step3: Determine the sign of (f^\prime(x))

We can test the intervals ((-\infty,\frac{16 - \sqrt{139}}{3})), ((\frac{16 - \sqrt{139}}{3},\frac{16+\sqrt{139}}{3})) and ((\frac{16+\sqrt{139}}{3},\infty)) using test - points. Let's take (x = 1) for the first interval: (f^\prime(1)=3\times1^{2}-32\times1 + 39=3-32 + 39 = 10>0). Let's take (x = 5) for the second interval: (f^\prime(5)=3\times5^{2}-32\times5 + 39=75-160 + 39=-46<0). Let's take (x = 10) for the third interval: (f^\prime(10)=3\times10^{2}-32\times10 + 39=300-320 + 39 = 19>0). So (f(x)) decreases on the interval ((\frac{16 - \sqrt{139}}{3},\frac{16+\sqrt{139}}{3})).

Step4: Calculate (a + b)

(a + b=\frac{16 - \sqrt{139}}{3}+\frac{16+\sqrt{139}}{3}=\frac{(16 - \sqrt{139})+(16+\sqrt{139})}{3}=\frac{32}{3}\approx10.667)

Answer:

(10.667)