High School

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Given the function f(x)=8x^(1/3), which of the following is a valid formula for the instantaneous rate of change at x=8?
1) lim_(h→0)(8h^(1/3)+16)/(h) lim
2) lim_(h→0)(8h^(1/3)-16)/(h) lim
3) lim_(h→0)(8h^(1/3)+8)/(h) lim
4) lim_(h→0)(8h^(1/3)-8)/(h) lim

Answer :

The correct option is:

3) [tex]\( \lim_{h \to 0} \frac{8h^{1/3} + 8}{h} \)[/tex]

How can you find this?

To find the instantaneous rate of change of a function at a given point, we can use the definition of the derivative. The derivative of a function gives us the slope of the tangent line to the curve at a specific point.

For the function [tex]\( f(x) = 8x^{1/3} \)[/tex], the derivative can be found using the power rule for differentiation.

[tex]\[ f'(x) = \frac{d}{dx} (8x^{1/3}) = \frac{8}{3}x^{-2/3} \][/tex]

Now, to find the instantaneous rate of change at x = 8 , we substitute x = 8 into the derivative:

[tex]\[ f'(8) = \frac{8}{3}(8)^{-2/3} = \frac{8}{3}(1/2) = \frac{4}{3} \][/tex]

So, the correct option is:

3) [tex]\( \lim_{h \to 0} \frac{8h^{1/3} + 8}{h} \)[/tex]

The complete question;

Given the function [tex]f(x)=8x^{(1/3)[/tex], which of the following is a valid formula for the instantaneous rate of change at x=8?

1) [tex]\( \lim_{h \to 0} \frac{8h^{1/3} + 16}{h} \)[/tex]

2) [tex]\( \lim_{h \to 0} \frac{8h^{1/3} - 16}{h} \)[/tex]

3) [tex]\( \lim_{h \to 0} \frac{8h^{1/3} + 8}{h} \)[/tex]

4) [tex]\( \lim_{h \to 0} \frac{8h^{1/3} - 8}{h} \)[/tex]

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Rewritten by : Barada