Find the inflection points of the function f(x)=4x^4-32x^3+89x^2-95x+23 and the interval where the function is concave up and concave down.    

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The function f(x)=4x^4 - 32x^3 + 89x^2 - 95x + 23

The derivative f'(x) = 16x^3 - 96x^2 + 178x - 95

f''(x) = 48*x^2-192*x+178

Solving 48*x^2 - 192*x + 178 = 0 gives the roots x = `(24 + sqrt 42)/12` and x = `(24 - sqrt 42)/12`

The inflection points lie at the points where x = `(24 + sqrt 42)/12 ` and x = `(24 - sqrt 42)/12`

This can be verified from the graph of the derivative of the function

The function is concave up in the interval where the f''(x)>=0 and concave down in the interval where f''(x) <= 0

f''(x) = `(x - (24 + sqrt 42)/12)(x - (24 - sqrt 42)/12)`

f''(x) `>=` 0 if `(x - (24 + sqrt 42)/12) >= 0` and (x` - (24 - sqrt 42)/12) >= 0` or if `(x - (24 + sqrt 42)/12) <= 0` and `(x - (24 - sqrt 42)/12) <= 0`

=> `x <= (24 - sqrt 42)/12` or `x >= (24 + sqrt 42)/12`

The function is concave down when it lies in `[(24 - sqrt 42)/12, (24 + sqrt 42)/12]`

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