Question

anyone who solves this problem for me .... I really need help

Answer 1

a) The value of $k'(0)$ is $\frac{3\sqrt{3}}{2}$.

b) The value of $m'(5)$ is approximately -0.034.

c) The value of $x$ is approximately 0.622.

a) $f(x)$ is a piecewise function formed by two linear functions, whose form is defined by the following definition:

$f(x) = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \cdot x + b$ (1)

Where:

• $(x_{1}, y_{1})$, $(x_{2}, y_{2})$ - Two distinct points of the line.
• $x$ - Independent variable.
• $f(x)$ - Dependent variable.
• $b$ - x-Intercept

Now we proceed to determine the linear functions:

Line 1: $x \in [-1, 2)$

$(x_{1}, y_{1}) = (0, 3)$, $(x_{2}, y_{2}) = (1, 5)$, $b = 3$

$f(x) = \frac{5-3}{1-0}\cdot x + 3$

$f(x) = 2\cdot x + 3$

Line 2: $x \in [2, 8]$

$(x_{1}, y_{1}) = (2, 7)$, $(x_{2}, y_{2}) = (8, 3)$

First, we determine the slope of function:

$m = \frac{3-7}{8-2}$

$m = -\frac{2}{3}$

Now we proceed to determine the intercept of the linear function:

$7 = -\frac{2}{3}\cdot 2 + b$

$b = \frac{25}{3}$

$f(x) = -\frac{2}{3}\cdot x +\frac{25}{3}$

The first derivative of a linear function is its slope, and the first derivative of a product of functions is defined by:

$k'(x) = f'(x)\cdot g(x) + f(x)\cdot g'(x)$

If we know that $f(x) = 2\cdot x + 3$, $f'(x) = 2$, $g(x) = \sqrt{x^{2}-x+3}$, $g'(x) = \frac{2\cdot x - 1}{2\cdot \sqrt{x^{2}-x+3}}$ and $x = 0$, then:

$k'(x) = 2\cdot \sqrt{x^{2}-x+3}+\frac{(2\cdot x +3)\cdot (2\cdot x - 1)}{2\cdot \sqrt{x^{2}-x+3}}$

$k'(0) = 2\sqrt{3}-\frac{3}{2\sqrt{3}}$

$k'(0) = \frac{12-3}{2\sqrt{3}}$

$k'(0) = \frac{9}{2\sqrt{3}}$

$k'(0) = \frac{3\sqrt{3}}{2}$

The value of $k'(0)$ is $\frac{3\sqrt{3}}{2}$.

b) The derivative is found by means of the formulas for the derivative of the product of a function and a constant and the derivative of a division between two functions:

$m'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{2\cdot [g(x)]^{2}}$ (2)

If we know that $f(x) = -\frac{2}{3}\cdot x +\frac{25}{3}$, $f'(x) = -\frac{2}{3}$, $g(x) = \sqrt{x^{2}-x+3}$, $g'(x) = \frac{2\cdot x - 1}{2\cdot \sqrt{x^{2}-x+3}}$ and $x = 5$, then:

$f(5) = 5$

$f'(5) = -\frac{2}{3}$

$g(5) = \sqrt{23}$

$g'(5) = \frac{9\sqrt{23}}{46}$

$m'(5) = \frac{\left(-\frac{2}{3} \right)\cdot \sqrt{23}-\left(-\frac{2}{3} \right)\left(\frac{9\sqrt{23}}{46} \right)}{3\cdot 25}$

$m'(5) \approx -0.034$

The value of $m'(5)$ is approximately -0.034.

c) In this case we must find a value of $x$, so that $h'(x) = 2$. Hence, we have the following formula below:

$5\cdot e^{x}-9\cdot \cos x = 2$

A quick approach is using a graphing tool a locate a point so that $5\cdot e^{x}-9\cdot \cos x = 2$. According to this, the value of $x$ is approximately 0.622.

To learn more on derivatives, we kindly invite to check this verified question: https://brainly.com/question/21202620