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9 mark for review the graph of the trigonometric function ( f ) is show…

Question

9 mark for review
the graph of the trigonometric function ( f ) is shown above for ( a leq x leq b ). at which of the following points
on the graph of ( f ) could the instantaneous rate of change of ( f ) equal the average rate of change of ( f ) on the
interval ( a, b )?

Explanation:

Step1: Recall the Mean Value Theorem

The Mean Value Theorem states that if a function \(y = f(x)\) is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), then there exists at least one number \(c\in(a,b)\) such that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\). Here, \(f^{\prime}(c)\) is the instantaneous rate of change of \(f\) at \(x = c\), and \(\frac{f(b)-f(a)}{b - a}\) is the average rate of change of \(f\) on the interval \([a,b]\).

Step2: Analyze the slope of the secant line and tangent line

The average rate of change \(\frac{f(b)-f(a)}{b - a}\) is the slope of the secant line connecting the points \((a,f(a))\) and \((b,f(b))\). The instantaneous rate of change \(f^{\prime}(x)\) is the slope of the tangent line to the curve \(y = f(x)\) at \(x\).
Looking at the graph:

  • At point \(A\), the slope of the tangent line (instantaneous rate of change) is \(0\) (horizontal tangent).
  • At point \(B\), the slope of the tangent line is negative.
  • At point \(C\), the slope of the tangent line (by visual inspection of the graph) has the same sign and magnitude as the slope of the secant line connecting \((a,f(a))\) and \((b,f(b))\) (since the secant line has a negative slope and the tangent line at \(C\) also has a negative slope and the "steepness" (magnitude) of the tangent at \(C\) matches the "steepness" of the secant line).
  • At point \(D\), the slope of the tangent line (instantaneous rate of change) is \(0\) (horizontal tangent).

Answer:

\(C\)