# Suppose f (x) is a function and f '(x) = -2x2 - x + 15, i.e. the DERIVATIVE of f (x) is -2x2 - x...

## Question:

Suppose {eq}f (x){/eq} is a function and {eq}f '(x) = -2x2 - x + 15{/eq}, i.e. the DERIVATIVE of {eq}f (x){/eq} is {eq}-2x2 - x + 15{/eq}. USE

BASIC ALGEBRA to find the {eq}x{/eq} coordinate of each critical point of {eq}f (x){/eq}. Write down the equation you solve to

find the critical points and SHOW all steps to solve it.

## Critical Points:

Since, the derivative of the given function is a quadratic equation and we know that to find the {eq}x {/eq} value of the critical points, we have to set the derivative of the function to zero. So here, we will set quadratic equation to zero. For solving this quadratic equation, we will use quadratic factorization using splitting mid terms.

The derivative of function is given by:

{eq}f^{'} (x) =- 2x^2 - x + 15 {/eq}

For finding critical points we have to set {eq}f^{'}(x) = 0 {/eq}

Hence:

{eq}f^{'}(x) = 0 \\ \Rightarrow -2x^2 - x + 15 = 0 \\ \Rightarrow - (2x^2 + x -15) = 0 \\ \Rightarrow 2x^2 + x - 15 = 0 \\ \Rightarrow 2x^2 + 6x - 5x - 15 = 0 \\ \Rightarrow 2x(x+ 3) -5(x + 3) = 0 \\ \Rightarrow (x+3)(2x - 5) = 0 {/eq}

So:

{eq}x + 3 = 0 \\ \Rightarrow x = -3 {/eq}

And:

{eq}2x - 5 = 0 \\ \Rightarrow x = \frac{5}{2} {/eq}

Hence:

The critical points are {eq}x = -3 \,\, and \,\, x = \frac{5}{2} {/eq}