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3 years ago

can anyone please help me I am struggling?? It’s 9th grade algebra 1 average rate of change comparison

Mathematics

1 answer:

gtnhenbr [62]3 years ago

5 0

Answer:

a) 2 b) 1 c) 7/6

Function a(x) = 2x + 3 has the greatest rate of change

Step-by-step explanation:

a(x) = 2x + 3

2(-1) + 3 = 1 (-1,1)

2(2) + 3 = 7 (2,7)

$\frac{7-1}{2-(-1)}$ = 6/3

b(x) = x^2 - 1

(-1)^2 - 1 = 0 (-1, 0)

(2)^2 - 1 = 3 (2,3)

$\frac{3-0}{2-(-1)}$ = 3/3

c(x) = 2^x + 1

2^(-1) + 1 = 1.5 (-1, 1.5)

2^(2) + 1 = 5 (2,5)

$\frac{5-1.5}{2-(-1)}$ = 3.5/3 or 7/6

=7/6

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Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y, z) = xe^y + ye^z +

LUCKY_DIMON [66]

Answer:

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}$

Step-by-step explanation:

We need to find the directional derivative of the function at the given point in the direction of the vector v.

$f(x, y, z)=xe^{y} + ye^{z} + ze^{x}$ ,point (0, 0, 0) and $v=$

By Theorem: If f is a differentiable function of x , y and z , then f has a directional derivative for any unit vector $\overrightarrow{v} =$ and

$D_{\overrightarrow{u}}f(x,y,z)=f_{x}(x,y,z)u_1+f_{y}(x,y,z)u_2+f_{z}(x,y,z)u_3$

where $\overrightarrow{u}=\frac{\overrightarrow{v}}{||v||}$

since, $v=$

then $\overrightarrow{u}=\frac{\overrightarrow{v}}{||v||}$

$\overrightarrow{u}=< \frac{6}{\sqrt{6^{2}+3^{2}+(-3)^{2}}},\frac{3}{\sqrt{6^{2}+3^{2}+(-3)^{2}}},\frac{-3}{\sqrt{6^{2}+3^{2}+(-3)^{2}}} >$

$\overrightarrow{u}=< \frac{6}{\sqrt{54}},\frac{3}{\sqrt{54}},\frac{-3}{\sqrt{54}} >$

The partial derivatives are

$f_{x}(x,y,z)=e^{y}+ze^{x}$

$f_{y}(x,y,z)=xe^{y}+e^{z}$

$f_{z}(x,y,z)=ye^{z}+e^{x}$

Then the directional derivative is

$D_{\vec{u}}f(x,y,z)=(e^{y}+ze^{x})(\frac{6}{\sqrt{54}})+(xe^{y}+e^{z})(\frac{3}{\sqrt{54}})+(ye^{z}+e^{x})(\frac{-3}{\sqrt{54}})$

so, directional derivative at point (0,0,0)

$D_{\vec{u}}f(0,0,0)=(e^{0}+0e^{0})(\frac{6}{\sqrt{54}})+(0e^{0}+e^{0})(\frac{3}{\sqrt{54}})+(0e^{0}+e^{0})(\frac{-3}{\sqrt{54}})$

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}+\frac{3}{\sqrt{54}}+\frac{-3}{\sqrt{54}}$

$D_{\vec{u}}f(0,0,0)=\frac{6+3-3}{\sqrt{54}}$

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}$

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4 years ago

(7x^2-2x+5) - (3x^2+4x+10) simplify and show steps

mixas84 [53]

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