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4 years ago

Let f(x) = 1850(0.96)^x b) Solve the equation f(x+1) = f(x) -25

Mathematics

1 answer:

Anton [14]4 years ago

6 0

Answer:

(a) -69.59 . . . . the approximate slope of the curve at x=2

(b) 26.583 . . . . the value of x where the slope of f(x) is about -25.

Step-by-step explanation:

(a) The slope of a function <em>f </em>over the interval [x, x+h] is given by ...

... slope = (f(x+h) -f(x))/h

The problem statement asks for the value of ...

... (f(2.01) -f(2))/0.01

Matching this expression to the pattern above, we find x = 2, h = 0.01. An appropriate interpretation of the expression is that it gives the approximate slope of f(x) near x=2.

The value of that slope is about ...

... (f(2.01) -f(2))/0.01 ≈ 1850·(0.92122386 -0.9216)/.01 ≈ -69.5857

(b) f(x+1) = 1850·0.96^(x+1) = 0.96·f(x)

Then the equation becomes ...

... 0.96·f(x) = f(x) -25

... 25 = 0.04·f(x) . . . . . . . . add 25-0.96·f(x)

... 625 = f(x) . . . . . . . . . . . multiply by 25

... 625 = 1850·0.96^x . . . substitute definition of f(x)

... 625/1850 = 0.96^x . . . divide by the coefficient of the exponential term

... log(625/1850) = x·log(0.96) . . . . take logs

... x = log(625/1850)/log(0.96) ≈ -0.471292/-0.0177288

... x ≈ 26.583

If you rearrange the equation to ...

... f(x+1) -f(x) = -25

Comparing this to the equation for slope, above, it looks a lot like we're trying to find the interval [x, x+1] over which the average slope is -25.

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n is a positive integer with exactly two different divisors greater than 1, how many positive factors does n^2 have? A. 4 B. 5 C

Evgesh-ka [11]

Answer:

C. 5

Step-by-step explanation:

Remember that for any integer $n$ , the integers $1 \text{and} n$ are both divisors (or factors) of $n$ . First, we will prove that n is a square, and then we will compute the factors of the n².

In this case, the integer n has exactly two different divisors greater than 1. It's impossible that $n=1$ , since 1 doesn't have positive factors greater than 1. Then $n>1$ , therefore $n$ itself is one of the required divisors. Denote by $a$ the other divisor greater than 1, and note that to satisfy the condition on the divisors, $a$ .

Because a divides n, there exists some integer $k$ such that $n=ak$ . We must have that $k>1$ , if not, then $k\leq 1$ , which implies that $ak=n\leq a$ , which contradicts the part above.

Now, $k>1$ and, by definition of divisibility, k divides n. Then k must be equal either to n or a, since we can't have three different divisors of this kind. If $k=n$ then $n=an$ and by cancellation, $1=a$ which is a contradiction. Therefore $k=a$ and $n=a^2$ .

We have that $n=(a^2)^2=a^4$ . We can write n as $n=a^3a=a^2a^2=a^4 \cdot 1$ . From the first equation, a divides n and a³ divides n. From the second equation, a² divides, and from the last one, 1 divides n and a⁴=n divides n. Thus n has exactly 5 positive factors.