EXPLANATION :
a. The instantaneous rate of change is the same as the slope of the curve at a given point.
The slope function of f(x) is the first derivative of that function, f'(x)
From the problem, we have :
[tex]g(x)=x^4-2x^3+1[/tex]
Calculate the first derivative :
[tex]\begin{gathered} g(x)=x^4-2x^3+1 \\ g^{\prime}(x)=4(x^{4-1})-3(2x^{3-1})+0 \\ g^{\prime}(x)=4x^3-6x^2 \end{gathered}[/tex]
To find the solve at x = 1, evaluate g'(x) at x = 1
[tex]\begin{gathered} g^{\prime}(1)=4(1)^3-6(1)^2 \\ g^{\prime}(1)=-2 \end{gathered}[/tex]
The instantaneous rate of change is -2
b. The instantaneous rate of change is constant at any x-value if the function is linear.
Since the given function is NOT a linear, then the instantaneous rate of change is NOT the same.
