Question

Let 'a' be a real number such that the function \(f(x)=ax^2+6x-15,\ x\in\mathbb{R}\) is increasing in
\(\left(-\infty,\frac34\right)\)
and decreasing in
\(\left(\frac34,\infty\right)\)
. Then the function \(g(x)=ax^2-6x+15,\ x\in\mathbb{R}\) has a :

• A. local maximum at \(x=-\frac34\)
• B. local minimum at \(x=-\frac34\)
• C. local maximum at \(x=\frac34\)
• D. local minimum at \(x=\frac34\)

Hint:

\(f\) changes from increasing to decreasing at \(x=\frac34\), so \(f^{\prime}\left(\frac34\right)=0\) and \(a\lt0\); use this to find \(a\), then locate the extremum of \(g\).

Answer 1

The Correct Option is A

\(f^{\prime}(x)=2ax+6\). \(f\) increases then decreases at \(x=\frac34\), so \(f^{\prime}\left(\frac34\right)=0\): \(\frac{3a}{2}+6=0\Rightarrow a=-4\) (negative, consistent with a maximum of \(f\)).
\(g(x)=-4x^2-6x+15\), \(g^{\prime}(x)=-8x-6=0\Rightarrow x=-\frac34\).
\(g^{\prime\prime}(x)=-8\lt0\), so \(g\) has a local maximum at \(x=-\frac34\).
Answer: (1) local maximum at \(x=-\frac34\)