For the following equation, what is the instantaneous rate of change at x=1 ? f(x)=-2x^(5)+2x^(2)-5 Answer:

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For the following equation, what is the instantaneous rate of change at  x=1 ?  f(x)=-2x^(5)+2x^(2)-5 Answer:

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Identify function and point: Identify the function and the point at which the instantaneous rate of change is required. The function given is f(x) = -2x^(5) + 2x^(2) - 5, and we need to find the instantaneous rate of change at x = 1.Calculate derivative of function: Calculate the derivative of the function f(x). The derivative of f(x) with respect to x is f'(x) = d/dx(-2x^(5)) + d/dx(2x^(2)) - d/dx(5). Using the power rule, the derivative of -2x^(5) is -10x^(4), the derivative of 2x^(2) is 4x, and the derivative of a constant is 0. So, f'(x) = -10x^(4) + 4x.Evaluate derivative at x=1: Evaluate the derivative at x = 1 to find the instantaneous rate of change at that point. Substitute x = 1 into the derivative f'(x) to get f'(1) = -10(1)^(4) + 4(1). This simplifies to f'(1) = -10 + 4.Calculate final instantaneous rate: Calculate the final value of the instantaneous rate of change at x = 1. f'(1) = -10 + 4 = -6.

For the following equation, what is the instantaneous rate of change at $x=1$ ? $\newline$ $f(x)=-2 x^{5}+2 x^{2}-5$ $\newline$ Answer:

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For the following equation, what is the instantaneous rate of change at x=1 x=1 x=1 ?\newlinef(x)=−2x5+2x2−5 f(x)=-2 x^{5}+2 x^{2}-5 f(x)=−2x5+2x2−5\newlineAnswer:

For the following equation, what is the instantaneous rate of change at $x=1$ ? $\newline$ $f(x)=-2 x^{5}+2 x^{2}-5$ $\newline$ Answer: