# Graphing question(calculus) Question is below. Graph the function with a local maximum at (-1,2), a local minimum at (1,-2), an absolute maximum at... |

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You need to remember that the critical values of the function represent the roots of first derivative of function. Since the function has 3 critical points, then the derivative has four roots and this means that you may consider the derivative a four order polynomial such that:

`f'(x) = ax^4 + bx^3 + cx^2 + dx  + e`

Hence, `f'(-1)=0 =gt a-b+c-d+e=0`

`f'(1)=0 =gt a+b+c+d+e=0`

`f'(1)+f'(-1)=0 =gt a+c+e=0`

`f'(3) = 0 =gt81a + 27b + 9c + 3d + e = 0`

`f'(-3) = 0 =gt 81a- 27b + 9c- 3d + e = 0`

`f'(3)+f'(-3)=0 =gt 162a + 18c + 2e = 0`

`81a + 9c + e = 0`

`81a + 9c = a + c =gt 80a = -8c =gt c=-10a`

You also know that the function f(x) may be found integrating f'(x) such that:

`int f'(x) dx = int (ax^4 + bx^3 + cx^2 + dx + e) dx`

`f(x) = ax^5/5 + bx^4/4 + cx^3/3 + dx^2/2 + ex + f`

`f(1)=-2 =gt a/5+b/4+c/3+d/2+e+f=-2`

`f(-1)=2 =gt -a/5+b/4-c/3+d/2-e+f= 2`

Adding `f(1) + f(-1) = b/2 + d + 2f = 0`

`f(3) = 18 =gt 243a/5 + 81b/4 + 27c/3 + 9d/2 + 3e + f = 18`

`f(-3) = -18 =gt -243a/5 + 81b/4 - 27c/3 + 9d/2- 3e + f = -18`

Adding `f(3) + f(-3) = 81b/2 + 9d + 2f = 0`

Thus, `b/2 + d = 81b/2 + 9d =gt -40b = 8d =gt d = -5b`

The function increases over (-3,-1), it has a local maximum at (-1,2), it decreases over (-1,1), it has a local minimum at (1,-2), it increases again over (1,3), it has an absolute maximum at (3,18).

Hence, evaluating the domain of the function yields R and analyzing the monotony of function, it follows this path:function increases over (-3,-1), it has a local maximum at (-1,2), it decreases over (-1,1), it has a local minimum at (1,-2), it increases again over (1,3), it has an absolute maximum at (3,18).

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