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For the following equation, what is the instantaneous rate of change at 
x=-2?

f(x)=-x^(3)-2x
Answer:

For the following equation, what is the instantaneous rate of change at x=2? x=-2 ? \newlinef(x)=x32x f(x)=-x^{3}-2 x \newlineAnswer:

Full solution

Q. For the following equation, what is the instantaneous rate of change at x=2? x=-2 ? \newlinef(x)=x32x f(x)=-x^{3}-2 x \newlineAnswer:
  1. Calculate Derivative: To find the instantaneous rate of change of the function at a specific point, we need to calculate the derivative of the function. The derivative of a function at a point gives us the slope of the tangent line at that point, which is the instantaneous rate of change.
  2. Apply Power Rule: The function given is f(x)=x32xf(x) = -x^3 - 2x. We will find the derivative f(x)f'(x) using the power rule. The power rule states that the derivative of xnx^n is nx(n1)n\cdot x^{(n-1)}.
  3. Evaluate at x=2x = -2: Applying the power rule to each term in the function:\newlineThe derivative of x3-x^3 is 3x2-3x^2 (using the power rule).\newlineThe derivative of 2x-2x is 2-2 (since the derivative of xx is 11 and the constant multiple rule applies).\newlineSo, f(x)=3x22f'(x) = -3x^2 - 2.
  4. Substitute and Calculate: Now we need to evaluate the derivative at x=2x = -2 to find the instantaneous rate of change at that point.\newlineSubstitute x=2x = -2 into f(x)f'(x):\newlinef(2)=3(2)22f'(-2) = -3(-2)^2 - 2.
  5. Substitute and Calculate: Now we need to evaluate the derivative at x=2x = -2 to find the instantaneous rate of change at that point.\newlineSubstitute x=2x = -2 into f(x)f'(x):\newlinef(2)=3(2)22f'(-2) = -3(-2)^2 - 2.Calculate the value:\newlinef(2)=3(4)2=122=14f'(-2) = -3(4) - 2 = -12 - 2 = -14.

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