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2 years ago

10

Suppose it is known that for a given differentiable function y=f(x), its tangent line (local linearization) at the point where a

=−4 is given by T(x)=5−7(x+4). What must be the values of f(−4) and f′(−4)?

1 answer:

AleksandrR [38]2 years ago

6 0

Answer:

y(-4) = 5

y'(-4) = -7

Step-by-step explanation:

Hi!

Since the tangent line T and the curve y must coincide at x=-4

y(-4) = T(-4) = 5

On the other hand, the derivative of the curve evaluated at -4 y'(x=-4) must be the slope of the tangent line. Which inspecting the tangent line T(x) is -7

That is:

y'(-4) = -7

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The monthly rate is 10%

Step-by-step explanation:

Given

Let

$x \to months$

$y \to mice$

So, we have:

$(x_1,y_1) = (0,120)$ --- Monday

$(x_2,y_2) = (1,132)$ --- One month later

Required

The function

The function is represented as:

$y=ab^x$

In $(x_1,y_1) = (0,120)$ , we have:

$120 = a * b^0$

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$120 = a$

$a=120$

In $(x_2,y_2) = (1,132)$ , we have:

$132 = a * b^1$

$132 = a * b$

Substitute: $a=120$

$132 = 120 * b$

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$b = \frac{132}{120}$

$b = 1.1$

So, the function is:

$y=ab^x$

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$y =a(1 + r)^x$

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$1 + r = b$

Make r the subject

$r = b-1$

Substitute $b = 1.1$

$r = 1.1-1$

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Express as percentage

$r = 0.1*100\%$

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