# Find the critical points a)h(x)=2x^3-23x^2-16x+1 b)h(x)=sqrt(1-7*x^2)

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### 1 Answer

a) You need to find the critical points of the given function `h(x) =` `2x^3-23x^2-16x+1` , hence, you should find the roots of the equation `h'(x) = 0` such that:

`h'(x) = (2x^3-23x^2-16x+1) => h'(x) = 6x^2 - 46x - 16`

You need to solve `h'(x) = 0` such that:

`6x^2 - 46x - 16 = 0 => 3x^2 - 23x - 8 = 0`

Using quadratic formula yields:

`x_(1,2) = (23+-sqrt(23^2 - 4*3*(-8)))/(2*3)`

`x_(1,2) = (23+-sqrt(529+96))/6 => x_(1,2) = (23+-sqrt625)/6`

`x_(1,2) = (23+-25)/6 => x_1 = 48/6 => x_1 = 8`

`x_2 = -2/6 => x_2 = -1/3`

Hence, evaluating the critical values of the function yields `x_1 = 8` and `x_2 = -1/3.`

You need to evaluate the crititcal points, hence, you need to substitute 8 and -`1/3` for x in h(x) such that:

`h(8) = 1024-184-128+1 = 713`

`h(-1/3) = -2/27 - 23/9 + 16/3 + 1`

`h(-1/3) = (-2 - 69 + 144 + 27)/27 => h(-1/3) = 100/27`

**Hence, evaluating the critical points of the function yields `(8,713)` and `(-1/3,100/27).` **

b) You need to find the critical points of the given function `h(x)=sqrt(1-7x^2)` , hence, you need to solve `h'(x) = 0` such that:

`h'(x) = -14x/(2sqrt(1-7x^2)) =gt h'(x) = -7x/(sqrt(1-7x^2))`

Since `sqrt(1-7x^2)!=0` , then `-7x = 0 =gt x = 0.`

You need to evaluate the critical point of the function, hence, you need to evaluate h(0) such that:

`h(0) = sqrt(1 - 7*0) => h(0) = sqrt1 => h(0) = 1`

**Hence, evaluating the critical point of the function yields `(0,1).` **