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3 years ago

Let g(x)=9x−10 and evaluate g(x+h)−g(x)/h

Mathematics

1 answer:

castortr0y [4]3 years ago

3 0

Answer:

$\frac{g(x+h)-g(x)}{h}=9$

Step-by-step explanation:

we have

$g(x)=9x-10$

To find out g(x+h) substitute the variable x by the variable (x+h) in the function g(x)

$g(x+h)=9(x+h)-10$

$g(x+h)=9x+9h-10$

Evaluate

$\frac{g(x+h)-g(x)}{h}$

we have

$g(x+h)=9x+9h-10$

$g(x)=9x-10$

substitute in the expression

$\frac{9x+9h-10-(9x-10)}{h}$

$\frac{9x+9h-10-9x+10)}{h}$

$\frac{9h}{h}$

$9$

therefore

$\frac{g(x+h)-g(x)}{h}=9$

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Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co

Olenka [21]

Answer:

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\$

Step-by-step explanation:

First, notice that:

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\$

We proceed to use the chain rule to find $g_{r}(\sqrt{2},\frac{\pi}{4})$ using the fact that $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ to find their derivatives:

$g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\$

Because we know $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ then:

$\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\delta y}{\delta r}=sin(\theta)$

We substitute in what we had:

$g_{r}(r,\theta)=f_{x}( rcos(\theta),rsin(\theta))cos(\theta)+f_{y}(rcos(\theta),rsin(\theta))sin(\theta)$

Now we put in the values $r=\sqrt{2}\ and\ \theta=\frac{\pi}{4}$ in the formula:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})$

Because of what we supposed:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})$

And we operate to discover that:

$g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}$

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

and this will be our answer

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2 years ago

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