Math

Question Use the second derivative test to find local extrema of f(x)=5x33+10x2+25xf(x) = -\frac{5x^3}{3} + 10x^2 + 25x in (5,8)(-5,8). The local maxima occur at x=105x = \frac{10}{5}. The local minima occur at \varnothing.

Studdy Solution

STEP 1

1. The function f(x)=5x33+10x2+25xf(x)=-\frac{5x^3}{3}+10x^2+25x is a polynomial function, which is differentiable everywhere.
2. The Second Derivative Test states that if f(a)>0f''(a) > 0, then ff has a local minimum at x=ax=a, and if f(a)<0f''(a) < 0, then ff has a local maximum at x=ax=a.
3. To apply the Second Derivative Test, we need to find the critical points where f(x)=0f'(x) = 0 and then evaluate the second derivative at those points.
4. The interval (5,8)(-5,8) is the domain of interest for finding local extrema.

STEP 2

1. Find the first derivative of the function f(x)f(x).
2. Determine the critical points by solving f(x)=0f'(x) = 0.
3. Find the second derivative of the function f(x)f(x).
4. Apply the Second Derivative Test to each critical point to determine if it is a local maximum or minimum.
5. Verify that the critical points are within the interval (5,8)(-5,8).

STEP 3

Find the first derivative of the function f(x)f(x).
f(x)=ddx(5x33+10x2+25x) f'(x) = \frac{d}{dx}\left(-\frac{5x^3}{3}+10x^2+25x\right)

STEP 4

Use the power rule to differentiate each term.
f(x)=5x2+20x+25 f'(x) = -5x^2 + 20x + 25

STEP 5

Find the critical points by solving f(x)=0f'(x) = 0.
5x2+20x+25=0 -5x^2 + 20x + 25 = 0

STEP 6

Factor the quadratic equation to solve for xx.
5(x24x5)=0 -5(x^2 - 4x - 5) = 0

STEP 7

Use the quadratic formula to find the roots of x24x5x^2 - 4x - 5.
x=(4)±(4)24(1)(5)2(1) x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}

STEP 8

Simplify the expression to find the values of xx.
x=4±16+202 x = \frac{4 \pm \sqrt{16 + 20}}{2}

STEP 9

Calculate the discriminant and the roots.
x=4±362 x = \frac{4 \pm \sqrt{36}}{2} x=4±62 x = \frac{4 \pm 6}{2}

STEP 10

Find the two critical points.
x=5andx=1 x = 5 \quad \text{and} \quad x = -1

STEP 11

Find the second derivative of the function f(x)f(x).
f(x)=ddx(5x2+20x+25) f''(x) = \frac{d}{dx}(-5x^2 + 20x + 25)

STEP 12

Differentiate each term using the power rule.
f(x)=10x+20 f''(x) = -10x + 20

STEP 13

Apply the Second Derivative Test to the critical point x=5x=5.
f(5)=10(5)+20 f''(5) = -10(5) + 20

STEP 14

Calculate the value of the second derivative at x=5x=5.
f(5)=50+20 f''(5) = -50 + 20 f(5)=30 f''(5) = -30

STEP 15

Since f(5)<0f''(5) < 0, ff has a local maximum at x=5x=5.

STEP 16

Apply the Second Derivative Test to the critical point x=1x=-1.
f(1)=10(1)+20 f''(-1) = -10(-1) + 20

STEP 17

Calculate the value of the second derivative at x=1x=-1.
f(1)=10+20 f''(-1) = 10 + 20 f(1)=30 f''(-1) = 30

STEP 18

Since f(1)>0f''(-1) > 0, ff has a local minimum at x=1x=-1.

STEP 19

Verify that the critical points x=5x=5 and x=1x=-1 are within the interval (5,8)(-5,8).
Both 55 and 1-1 are within the interval (5,8)(-5,8).
The local maxima occur at x=5x=5. The local minima occur at x=1x=-1.

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