Can someone please help me with this one question?

Mathematics

1 answer:

Scilla [17]3 years ago

8 0

It's been awhile since I've worked with infinite series and partial sums, but here are my thoughts:
A. the first 4 terms: -4, $- \frac{4}{3}$ , $-\frac{4}{9}$ , $- \frac{4}{27}$
B. In examining the partial sums of the series, it appears that the sequence produced has a limit and therefore the series would be convergent.
C. the limit or sum of the series looks to be -5 Below look at the sequence of partial sums:
-4, $-5 \frac{1}{3}$ , $-5 \frac{7}{9}$ , $-5 \frac{25}{27}$

So you can see we're getting closer and closer to -5

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Which value, to the nearest tenth, is the smallest solution of f(x) = g(x) if f(x) = ln(x + 5) – 1 and g(x) = x3 – 2x + 1?

nekit [7.7K]

Answer:

There are two possibilities of determine which solution is the smallest one:

(i) The smallest solution is the value of $x$ has the smallest distance with respect to origin: $(x_{2},y_{2}) = (0.18,0)$

(ii) The smallest solution is the most negative value: $(x_{1},y_{1}) = (-1.577,0)$

Step-by-step explanation:

Let $f(x) = g(x)$ , $f(x) = \ln (x+5)-1$ and $g(x) = x^{3}-2\cdot x + 1$ , then we have the following implicit equation:

$\ln (x+5)-x^{3}+2\cdot x -2 = 0$ (1)

This equation cannot be solved by analytical means. There are two different strategies to solve this expression: (i) Numerical methods, (ii) Graphical methods. We decide to use the second method:

Let suppose that we use this function $h(x) = \ln (x+5) -x^{3}+2\cdot x - 2$ . By the help of a graphing tool, we find the smallest solution such that $h(x) = 0$ . There are three possible solutions: $(x_{1},y_{1}) = (-1.577,0)$ , $(x_{2},y_{2}) = (0.18,0)$ , $(x_{3}, y_{3}) = (1.376, 0)$