Question

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Answer 1

$f(x) = mx^2-8x-3 \\ \\ f'(x) = 2mx-8 \\ f'(x) = 0 \Rightarrow 2mx = 8 \Rightarrow x = \dfrac{4}{m}\\ \\ f_{max} = 5\Rightarrow m < 0 \\ \\ \Rightarrow f_{max} =f\Big(\frac{4}{m}\Big)=5 \Rightarrow \dfrac{16}{m}-\dfrac{32}{m}-3 = 5 \Rightarrow \\ \\ \Rightarrow 16-32 = 8m \Rightarrow \boxed{m = -2}$