What are the extrema of #f(x)=2x^3-3x^2-36x-3#?

Question

Ezekiel Alexander · Accepted Answer

Local maximam is at # (-2,41)# and local minimum
is at # (3,-84)# All local maximums and minimums on a function’s graph are called local extrema. #f(x) =2x^3-3x^2-36x-3 or f^'(x) = 6x^2 -6x- 36# At critical points #f^'(x)=0 :. 6(x^2-x-6)=0 # or #x^2-x-6=0 or (x-3)(x+2) =0 # ;Critical points are #x=-2, x=3 # when #x=-2 , f(x)= 2(-2)^3-3(-2)^2-36(-2)-3=41# when #x=3 , f(x)= 2* 3^3-3*3^2-36*3-3= -84# . To test local maximum or minimum we will examine sign change at three positions #x = -3 , x =0 ,x=4 # for #f^'(x) = 6x^2 -6x- 36 ; f^'( -3) = 36 # (increasing) , #f^'(0) = -36 # (decreasing) and #f^'(4) = 36 # (increasing) #x=-2# is local maximum and #x=3 # is local minimum Hence local maximam is at # (-2,41)# and local minimum is at # (3,-84)# graph{2x^3-3x^2-36x-3 [-157, 157, -78.5, 78.5]}

Cameron Alexander · Answer

undefined To find the extrema of the function $ f(x) = 2x^3 - 3x^2 - 36x - 3 $, we first need to find its critical points by taking the derivative and setting it equal to zero. Then, we determine whether these critical points correspond to local minima or maxima by examining the sign of the second derivative. 1. Find the derivative of $ f(x) $: $$ f'(x) = 6x^2 - 6x - 36 $$ 2. Set the derivative equal to zero and solve for $ x $: $$ 6x^2 - 6x - 36 = 0 $$ $$ x^2 - x - 6 = 0 $$ $$ (x - 3)(x + 2) = 0 $$ 3. Find the critical points: $$ x = 3 \quad \text{and} \quad x = -2 $$ 4. Find the second derivative of $ f(x) $: $$ f''(x) = 12x - 6 $$ 5. Evaluate the second derivative at the critical points: $$ f''(3) = 12(3) - 6 = 30 > 0 $$ $$ f''(-2) = 12(-2) - 6 = -30 < 0 $$ 6. Analyze the sign of the second derivative: - $ f''(3) > 0 $: This indicates a local minimum at $ x = 3 $. - $ f''(-2) < 0 $: This indicates a local maximum at $ x = -2 $. Thus, the function $ f(x) = 2x^3 - 3x^2 - 36x - 3 $ has a local minimum at $ x = 3 $ and a local maximum at $ x = -2 $.

Elijah Abram · Answer

undefined To find the extrema of the function $ f(x) = 2x^3 - 3x^2 - 36x - 3 $, we first need to find its critical points. We do this by finding the first derivative of the function, setting it equal to zero, and solving for $ x $.

The first derivative of $ f(x) $ is $ f'(x) = 6x^2 - 6x - 36 $.

Setting $ f'(x) $ equal to zero and solving for $ x $, we get:

$ 6x^2 - 6x - 36 = 0 $

Dividing both sides by 6, we get:

$ x^2 - x - 6 = 0 $

This is a quadratic equation that can be factored as:

$ (x - 3)(x + 2) = 0 $

So, the critical points are $ x = 3 $ and $ x = -2 $.

To determine if these critical points are relative maxima or minima, we use the second derivative test.

The second derivative of $ f(x) $ is $ f''(x) = 12x - 6 $.

Evaluating $ f''(3) $, we get:

$ f''(3) = 12(3) - 6 = 36 - 6 = 30 $

Since $ f''(3) > 0 $, the function has a relative minimum at $ x = 3 $.

Evaluating $ f''(-2) $, we get:

$ f''(-2) = 12(-2) - 6 = -24 - 6 = -30 $

Since $ f''(-2) < 0 $, the function has a relative maximum at $ x = -2 $.

Therefore, the relative extrema of $ f(x) = 2x^3 - 3x^2 - 36x - 3 $ are:

Relative minimum at $ x = 3 $ and relative maximum at $ x = -2 $.