# Practice Paper 1 Question 6

Which values of \(k\) give a maximum at \(x=-1\) for \(f(x)=(k+1)x^4-(3k+2)x^2-2kx\)?

## Related topics

## Hints

- Hint 1How do you find points of maxima/minima of a given curve?
- Hint 2What is the value of derivative of \(f(x)\) at \(x=-1\)?
- Hint 3How do you determine if a stationary point is a point of maxima?

## Solution

First derivative of \(f(x)\) with respect to \(x\) is \((4k+4)x^3-(6k+4)x-2k\), and second derivative of \(f(x)\) with respect to \(x\) is \((12k+12)x^2-(6k+4)\). Notice that the first derivative is always \(0\) at \(x=-1,\) it does not depend on \(k\). For maxima, the second derivative must be negative at \(x=-1\) which gives us \(12k+12 -(6k+4)<0,\) and hence \(k<-\frac{4}{3}\).

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