1. The graph of the function \( f \) is shown below.

The function \( h \) is defined by \( h(x) = f(2x^2 - x) \). Find the slope of the line tangent to the graph of \( h \) at the point where \( x = -1 \).






Answer is: option2

\( \frac{5}{2} \)

Solution:

.

\( h'(x) = f'(2x^2 - x) \cdot (4x - 1) \)

\( h'(-1) = f'(2(-1)^2 - (-1)) \cdot (4(-1) - 1) \)

\( h'(-1) = f'(3) \cdot (-5) \)

\( h'(-1) = \left( -\frac{1}{2} \right) \cdot (-5) = \frac{5}{2} \)

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