Question

webwork 4 - topics 8 - 9: problem 5
(1 point)
consider the function ( f(x)=-4 x^{2}+6 x - 10 ). find the critical point, ( a ), of the function.
( a=)
at ( x = a ), does ( f(x) ) have a local min, a local max, or neither? type in your answer as lmin, lmax, or neither.
answer

Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=-4x^{2}+6x - 10\) using the power rule \((x^{n})^\prime=nx^{n - 1}\) is \(f^\prime(x)=-8x + 6\).

Step2: Set the derivative equal to zero and solve for \(x\)

Set \(f^\prime(x)=0\), so \(-8x+6 = 0\).
Add \(8x\) to both sides: \(6=8x\).
Divide both sides by \(8\): \(x=\frac{6}{8}=\frac{3}{4}\).

Step3: Use the second - derivative test

Find the second derivative \(f^{\prime\prime}(x)\). Differentiating \(f^\prime(x)=-8x + 6\) gives \(f^{\prime\prime}(x)=-8\).
Since \(f^{\prime\prime}(\frac{3}{4})=-8<0\), the function has a local maximum at \(x = \frac{3}{4}\).

Answer:

\(A=\frac{3}{4}\)