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3 years ago

Find the derivative of the function using the definition of derivative.f(x) = mx + qf '(x) =

Mathematics

1 answer:

Natasha_Volkova [10]3 years ago

5 0

Answer:

a) $f'(x) = m$

b) $x \in (-\infty, \infty)$

c) $x \in (-\infty, \infty)$

Step-by-step explanation:

We are given the following in the question:

$f(x) = mx + q$

a) We have to find the derivative of the given function.

$f'(x) = \dfrac{f(x+h)-f(x)}{h}\\\\= \dfrac{m(x+h)+q - mx - q}{h}\\\\f'(x) = \dfrac{mh}{h}\\\\f'(x) = m$

b) Domain of f(x)

Domain is the collection of all values of x for which the function is defined.

Domain of f(x) is all real numbers.

$x \in (-\infty, \infty)$

c) Domain of f'(x)

Domain of f'(x) is all real numbers.

$x \in (-\infty, \infty)$

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Please help on this. I am confused.

neonofarm [45]

The answer is 2x^2-3x-2.

To solve this, you take f(x)-g(x) which is 2x^2-5-3x+3. This simplifies to 2x^2-3x-2.

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3 years ago

Engineers want to design seats in commercial aircraft so that they are wide enough to fit 9090% of all males. (Accommodating 1

Molodets [167]

Answer:

Hip breadths less than or equal to 16.1 in. includes 90% of the males.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 14.5

Standard Deviation, σ = 1.2

We are given that the distribution of hip breadths is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.10.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 14.5}{1.2})=0.10$

$= 1 -P( z \leq \displaystyle\frac{x - 14.5}{1.2})=0.10$

$=P( z \leq \displaystyle\frac{x - 14.5}{1.2})=0.90$

Calculation the value from standard normal z table, we have,

$P(z < 1.282) = 0.90$

$\displaystyle\frac{x - 14.5}{1.2} = 1.282\\x = 16.0384 \approx 16.1$

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3 years ago

Find the equation of direct variation,

zysi [14]

Answer G. 99/36= 11/4x

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2 years ago

For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f

butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

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3 years ago

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Sladkaya [172]

Answer:

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Step-by-step explanation:

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3 years ago