Test you Knowledge: Applications of Derivatives

1. Find the maximum and minimum of f(x) = x^3-3x+1 on [0,3]

First, we need to find the zeros of the derivative to determine where there are critical values.

f'(x) = 3x^2-3 = 3(x^2-1)
This has zeros at x = 1 and x = -1. Now, we just need to test the values.

f(0) = 1
f(1) = -1
f(-1) = 3
f(3) = 19

Therefore, there is a max at x = 3 and a min at x = 1

2. Find a number c that satisfies the mean value theorem for f(x) = 3x^2+2x+5 on [-1,1].

First, we assert that f(x) is differentiable and continous on [-1,1], since it is a polynomial. Therefore, we know there is a point c such that f'(c) = \frac{f(b)-f(a)}{b-a}

In this problem, we are given that b = -1 and a = 1. From here we can just plug the values into the equation and solve it.

f'(c) = \frac{f(-1) - f(1)}{-1-1} = \frac{(3-2+5)-(3+2+5)}{-2} = \frac{6-10}{-2} = \frac{-4}{-2} = 2

Now, we just need to find where f'(c) = 2.

2 = 6x+2 \Longrightarrow 6x = 0 \Longrightarrow x = 0

3. Each side of a square is increasing at a rate of 6 cm/s. At which rate is the area of the square increasing when the area is 16cm^2?

Let’s define the length of the square as x. We know that the area of a square is A = x * x.

We also know that each side of the square is increasing 6 cm/s. Therefore we know that \frac{dx}{dt} = 6cm/s

We want to know the rate of the area increase when the area is 16cm^2. To start, we know that the area is 16cm^2 when x = 4, since A = x*x.

So, we get the following formula: A = x^2 \LongrightArrow \frac{dA}{dt} = 2x * \frac{dx}{dt}

From here, we just need to put in what we know. Since \frac{dx}{dt} = 6 and x = 4, we get:

\frac{dA}{dt} = 2*4 *6 = 48

4. Solve lim_{x \to 0} \frac{e^x-1-x}{x^2}

Start by checking the value of the function at x = 0.

\frac{e^0 - 1 - 0}{0^2} = \frac{0}{0}

This is indeterminate form, so we can apply L’Hoptial’s rule:

lim_{x \to 0} \frac{e^x-1-x}{x^2} = lim_{x \to 0} \frac{e^x-1}{2x}

This is still indeterminate form, so we can apply L’Hopital’s rule again:

lim_{x \to 0} \frac{e^x-1}{2x} = lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}

5. Use Newton’s Method to estimate the root of x^4-2x^3+5x^2-6 on [1,2]

x_2 = 2 - \frac{f(2)}{f'(2)} = 2 - \frac{14}{28} = \frac{3}{2} = 1.5

x_3 = 1.5 - \frac{f(1.5)}{f'(1.5)} = 1.26

x_4 = 1.26 - \frac{f(1.26)}{f'(1.26)} = 1.22

From here we could keep calculating values, but it looks like the root trends to around 1.2, therefore x = 1.2 is a good estimate towards to root.

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