Find lim ?x approaches 0 f(x+?x)-f(x)/?x where f(x) = 4x-3

Mathematics

1 answer:

Whitepunk [10]3 years ago

7 0

If $f(x)=4x-3$ :

$\displaystyle\lim_{\Delta x\to0}\frac{(4(x+\Delta x)-3)-(4x-3)}{\Delta x}=\lim_{\Delta x\to0}\frac{4\Delta x}{\Delta x}=4$

If $f(x)=4x^{-3}$ :

$\displaystyle\lim_{\Delta x\to0}\frac{\frac4{(x+\Delta x)^3}-\frac4{x^3}}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac{4x^3-4(x+\Delta x)^3}{x^3(x+\Delta x)^3}}{\Delta x}$

$\displaystyle=\lim_{\Delta x\to0}\frac{4x^3-4(x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3)}{x^3\Delta x(x+\Delta x)^3}$

$\displaystyle=\lim_{\Delta x\to0}\frac{-12x^2\Delta x-12x(\Delta x)^2-4(\Delta x)^3}{x^3\Delta x(x+\Delta x)^3}=-\frac{12}{x^4}$

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astra-53 [7]

Answer:

D. $f(x)=x^2$ and $k(x)=x^2-5$

Step-by-step explanation:

The vertex of a function is controlled by the translations of the x and y value. Moving a vertex up and down is connected to the vertical position or the y-value. The y-value is the value that is added to the variable and is not squared.

Vertex form of a function is $f(x)=a(x-h)^2+k$ . In this case, k controls the vertical position of the vertex. If k is negative then the vertex will be moved down. So, if you want to move a vertex down by 5 units, then the k value must be -5.