Find the turning points of the following functions of x and also determine their nature:- Y=(3X-5)3(3X-7)please could you work it step by steps. thanks
You need to find derivative of the function `y = (3x-5)^3*(3x-7). ` Check if derivative changes its sign at a point. If the derivative changes its sign at that point, then you have found a turning point.
Use the product rule, power rule and chain rule to find the derivative.
`dy/dx = 3(3x-5)^2*(3x-7) + 3(3x-5)^3`
Expand the binomials:
`dy/dx = 3(9x^2 - 30x + 25)(3x-7) + 3(27x^3 - 135x^2 + 225x - 125)`
Opening the brackets yields:
`dy/dx = 81x^3 - 189x^2 - 270x^2 + 630x + 225x - 525 + 81x^3 - 405x^2 + 675x - 375`
`` `dy/dx = 162x^3 - 864x^2 + 1530x - 900`
`dy/dx = 9(18x^3 - 96x^2 + 170x - 100)`
Put `dy/dx = 0` => `18x^3 - 96x^2 + 170x - 100 =` 0
You should find the roots of the equation `18x^3 - 96x^2 + 170x - 100 = 0` among the divisors of 100/18.
The roots of `18x^3 - 96x^2 + 170x - 100 = 0` become the x coordinates of the turning points of the function.