Solved on Feb 01, 2024

Graph $f(x) = x^2(4 - 5x)^3$ and estimate any relative extrema.

STEP 1

Assumptions
1. The function to graph is $f(x) = x^{2}(4-5x)^{3}$ .
2. Relative extrema refer to the local minimums and maximums of the function.
3. To find the relative extrema, we need to find the critical points where the first derivative of the function is zero or undefined.
4. We will use calculus techniques to find the derivative of the function.

STEP 2

First, we need to find the first derivative of the function $f(x)$ . To do this, we will use the product rule which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.
$(fg)' = f'g + fg'$

STEP 3

Let's identify the two functions that are being multiplied. We have $f(x) = u(x)v(x)$ where $u(x) = x^{2}$ and $v(x) = (4-5x)^{3}$ .

STEP 4

We will now find the derivatives of $u(x)$ and $v(x)$ .
For $u(x) = x^{2}$ , the derivative is:
$u'(x) = 2x$

STEP 5

For $v(x) = (4-5x)^{3}$ , we will use the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
$(g(h(x)))' = g'(h(x)) \cdot h'(x)$

STEP 6

Let's identify the outer function $g(y) = y^{3}$ and the inner function $h(x) = 4-5x$ .

STEP 7

Now, find the derivative of the outer function $g(y)$ with respect to $y$ .
$g'(y) = 3y^{2}$

STEP 8

Next, find the derivative of the inner function $h(x)$ with respect to $x$ .
$h'(x) = -5$

STEP 9

Apply the chain rule to find $v'(x)$ .
$v'(x) = g'(h(x)) \cdot h'(x) = 3(4-5x)^{2} \cdot (-5)$

STEP 10

Now, we can find the first derivative of $f(x)$ using the product rule.
$f'(x) = u'(x)v(x) + u(x)v'(x)$

STEP 11

Substitute $u'(x)$ , $v(x)$ , and $v'(x)$ into the product rule formula.
$f'(x) = 2x(4-5x)^{3} + x^{2}(3(4-5x)^{2} \cdot (-5))$

STEP 12

Simplify the expression for $f'(x)$ .
$f'(x) = 2x(4-5x)^{3} - 15x^{2}(4-5x)^{2}$

STEP 13

To find the critical points, we need to solve for $x$ when $f'(x) = 0$ or when $f'(x)$ is undefined.

STEP 14

Notice that $f'(x)$ is a polynomial and thus is never undefined. We only need to find where $f'(x) = 0$ .

STEP 15

Set $f'(x)$ to zero and solve for $x$ .
$2x(4-5x)^{3} - 15x^{2}(4-5x)^{2} = 0$

STEP 16

Factor out the common terms.
$x(4-5x)^{2}(2(4-5x) - 15x) = 0$

STEP 17

Set each factor equal to zero to find the critical points.
$x = 0, \quad (4-5x)^{2} = 0, \quad 2(4-5x) - 15x = 0$

STEP 18

Solve for $x$ in each equation.
From $x = 0$ , we have one critical point at $x = 0$ .

STEP 19

Solve $(4-5x)^{2} = 0$ by taking the square root of both sides.
$4-5x = 0$
$5x = 4$
$x = \frac{4}{5}$
So we have another critical point at $x = \frac{4}{5}$ .

STEP 20

Solve $2(4-5x) - 15x = 0$ .
$8 - 10x - 15x = 0$
$-25x + 8 = 0$
$25x = 8$
$x = \frac{8}{25}$
So we have another critical point at $x = \frac{8}{25}$ .

STEP 21

Now we have all the critical points: $x = 0$ , $x = \frac{4}{5}$ , and $x = \frac{8}{25}$ .

STEP 22

To determine whether these critical points are relative maxima or minima, we can use the first derivative test or the second derivative test. For simplicity, we will use the first derivative test by checking the sign of $f'(x)$ around each critical point.

STEP 23

Create a sign chart for $f'(x)$ around the critical points. This involves picking test points between and around the critical points and substituting them into $f'(x)$ to check the sign.

STEP 24

Choose test points $x = -1$ , $x = \frac{1}{4}$ , $x = \frac{1}{2}$ , and $x = 1$ to check the sign of $f'(x)$ around the critical points.

STEP 25

Substitute each test point into $f'(x)$ to determine the sign.
For $x = -1$ , $f'(x)$ is positive. For $x = \frac{1}{4}$ , $f'(x)$ is negative. For $x = \frac{1}{2}$ , $f'(x)$ is positive. For $x = 1$ , $f'(x)$ is negative.

STEP 26

From the sign chart, we can see that $f'(x)$ changes from positive to negative at $x = \frac{8}{25}$ , indicating a relative maximum at this point.

STEP 27

$f'(x)$ changes from negative to positive at $x = 0$ , indicating a relative minimum at this point.

STEP 28

$f'(x)$ changes from positive to negative at $x = \frac{4}{5}$ , indicating a relative maximum at this point.

STEP 29

To graph the function, plot the critical points and sketch the curve according to the sign chart, showing the relative maximums and minimums.

STEP 30

The estimated relative extrema are a relative minimum at $x = 0$ and relative maximums at $x = \frac{8}{25}$ and $x = \frac{4}{5}$ . The graph would show a parabolic curve opening upwards with these extrema.

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Graph f(x)=x2(4−5x)3f(x) = x^2(4 - 5x)^3f(x)=x2(4−5x)3 and estimate any relative extrema.

STEP 1

STEP 2

STEP 3

Let's identify the two functions that are being multiplied. We have f(x)=u(x)v(x) f(x) = u(x)v(x) f(x)=u(x)v(x) where u(x)=x2 u(x) = x^{2} u(x)=x2 and v(x)=(4−5x)3 v(x) = (4-5x)^{3} v(x)=(4−5x)3.

STEP 4

We will now find the derivatives of u(x) u(x) u(x) and v(x) v(x) v(x).For u(x)=x2 u(x) = x^{2} u(x)=x2, the derivative is:u′(x)=2x u'(x) = 2x u′(x)=2x

STEP 5

STEP 6

Let's identify the outer function g(y)=y3 g(y) = y^{3} g(y)=y3 and the inner function h(x)=4−5x h(x) = 4-5x h(x)=4−5x.

STEP 7

Now, find the derivative of the outer function g(y) g(y) g(y) with respect to y y y.g′(y)=3y2 g'(y) = 3y^{2} g′(y)=3y2

STEP 8

Next, find the derivative of the inner function h(x) h(x) h(x) with respect to x x x.h′(x)=−5 h'(x) = -5 h′(x)=−5

STEP 9

Apply the chain rule to find v′(x) v'(x) v′(x).v′(x)=g′(h(x))⋅h′(x)=3(4−5x)2⋅(−5) v'(x) = g'(h(x)) \cdot h'(x) = 3(4-5x)^{2} \cdot (-5) v′(x)=g′(h(x))⋅h′(x)=3(4−5x)2⋅(−5)

STEP 10

Now, we can find the first derivative of f(x) f(x) f(x) using the product rule.f′(x)=u′(x)v(x)+u(x)v′(x) f'(x) = u'(x)v(x) + u(x)v'(x) f′(x)=u′(x)v(x)+u(x)v′(x)

STEP 11

Substitute u′(x) u'(x) u′(x), v(x) v(x) v(x), and v′(x) v'(x) v′(x) into the product rule formula.f′(x)=2x(4−5x)3+x2(3(4−5x)2⋅(−5)) f'(x) = 2x(4-5x)^{3} + x^{2}(3(4-5x)^{2} \cdot (-5)) f′(x)=2x(4−5x)3+x2(3(4−5x)2⋅(−5))

STEP 12

Simplify the expression for f′(x) f'(x) f′(x).f′(x)=2x(4−5x)3−15x2(4−5x)2 f'(x) = 2x(4-5x)^{3} - 15x^{2}(4-5x)^{2} f′(x)=2x(4−5x)3−15x2(4−5x)2

STEP 13

To find the critical points, we need to solve for x x x when f′(x)=0 f'(x) = 0 f′(x)=0 or when f′(x) f'(x) f′(x) is undefined.

STEP 14

Notice that f′(x) f'(x) f′(x) is a polynomial and thus is never undefined. We only need to find where f′(x)=0 f'(x) = 0 f′(x)=0.

STEP 15

Set f′(x) f'(x) f′(x) to zero and solve for x x x.2x(4−5x)3−15x2(4−5x)2=0 2x(4-5x)^{3} - 15x^{2}(4-5x)^{2} = 0 2x(4−5x)3−15x2(4−5x)2=0

STEP 16

Factor out the common terms.x(4−5x)2(2(4−5x)−15x)=0 x(4-5x)^{2}(2(4-5x) - 15x) = 0 x(4−5x)2(2(4−5x)−15x)=0

STEP 17

Set each factor equal to zero to find the critical points.x=0,(4−5x)2=0,2(4−5x)−15x=0 x = 0, \quad (4-5x)^{2} = 0, \quad 2(4-5x) - 15x = 0 x=0,(4−5x)2=0,2(4−5x)−15x=0

STEP 18

Solve for x x x in each equation.From x=0 x = 0 x=0, we have one critical point at x=0 x = 0 x=0.

STEP 19

Solve (4−5x)2=0 (4-5x)^{2} = 0 (4−5x)2=0 by taking the square root of both sides.4−5x=0 4-5x = 0 4−5x=05x=4 5x = 4 5x=4x=45 x = \frac{4}{5} x=54​So we have another critical point at x=45 x = \frac{4}{5} x=54​.

STEP 20

Solve 2(4−5x)−15x=0 2(4-5x) - 15x = 0 2(4−5x)−15x=0.8−10x−15x=0 8 - 10x - 15x = 0 8−10x−15x=0−25x+8=0 -25x + 8 = 0 −25x+8=025x=8 25x = 8 25x=8x=825 x = \frac{8}{25} x=258​So we have another critical point at x=825 x = \frac{8}{25} x=258​.

STEP 21

Now we have all the critical points: x=0 x = 0 x=0, x=45 x = \frac{4}{5} x=54​, and x=825 x = \frac{8}{25} x=258​.

STEP 22

To determine whether these critical points are relative maxima or minima, we can use the first derivative test or the second derivative test. For simplicity, we will use the first derivative test by checking the sign of f′(x) f'(x) f′(x) around each critical point.

STEP 23

Create a sign chart for f′(x) f'(x) f′(x) around the critical points. This involves picking test points between and around the critical points and substituting them into f′(x) f'(x) f′(x) to check the sign.

STEP 24

Choose test points x=−1 x = -1 x=−1, x=14 x = \frac{1}{4} x=41​, x=12 x = \frac{1}{2} x=21​, and x=1 x = 1 x=1 to check the sign of f′(x) f'(x) f′(x) around the critical points.

STEP 25

STEP 26

From the sign chart, we can see that f′(x) f'(x) f′(x) changes from positive to negative at x=825 x = \frac{8}{25} x=258​, indicating a relative maximum at this point.

STEP 27

f′(x) f'(x) f′(x) changes from negative to positive at x=0 x = 0 x=0, indicating a relative minimum at this point.

STEP 28

f′(x) f'(x) f′(x) changes from positive to negative at x=45 x = \frac{4}{5} x=54​, indicating a relative maximum at this point.

STEP 29

To graph the function, plot the critical points and sketch the curve according to the sign chart, showing the relative maximums and minimums.

STEP 30

The estimated relative extrema are a relative minimum at x=0 x = 0 x=0 and relative maximums at x=825 x = \frac{8}{25} x=258​ and x=45 x = \frac{4}{5} x=54​. The graph would show a parabolic curve opening upwards with these extrema.

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Graph $f(x) = x^2(4 - 5x)^3$ and estimate any relative extrema.

Let's identify the two functions that are being multiplied. We have $f(x) = u(x)v(x)$ where $u(x) = x^{2}$ and $v(x) = (4-5x)^{3}$ .

We will now find the derivatives of $u(x)$ and $v(x)$ .
For $u(x) = x^{2}$ , the derivative is:
$u'(x) = 2x$

Let's identify the outer function $g(y) = y^{3}$ and the inner function $h(x) = 4-5x$ .

Now, find the derivative of the outer function $g(y)$ with respect to $y$ .
$g'(y) = 3y^{2}$

Next, find the derivative of the inner function $h(x)$ with respect to $x$ .
$h'(x) = -5$

Apply the chain rule to find $v'(x)$ .
$v'(x) = g'(h(x)) \cdot h'(x) = 3(4-5x)^{2} \cdot (-5)$

Now, we can find the first derivative of $f(x)$ using the product rule.
$f'(x) = u'(x)v(x) + u(x)v'(x)$

Substitute $u'(x)$ , $v(x)$ , and $v'(x)$ into the product rule formula.
$f'(x) = 2x(4-5x)^{3} + x^{2}(3(4-5x)^{2} \cdot (-5))$

Simplify the expression for $f'(x)$ .
$f'(x) = 2x(4-5x)^{3} - 15x^{2}(4-5x)^{2}$

To find the critical points, we need to solve for $x$ when $f'(x) = 0$ or when $f'(x)$ is undefined.

Notice that $f'(x)$ is a polynomial and thus is never undefined. We only need to find where $f'(x) = 0$ .

Set $f'(x)$ to zero and solve for $x$ .
$2x(4-5x)^{3} - 15x^{2}(4-5x)^{2} = 0$

Factor out the common terms.
$x(4-5x)^{2}(2(4-5x) - 15x) = 0$

Set each factor equal to zero to find the critical points.
$x = 0, \quad (4-5x)^{2} = 0, \quad 2(4-5x) - 15x = 0$

Solve for $x$ in each equation.
From $x = 0$ , we have one critical point at $x = 0$ .

Solve $(4-5x)^{2} = 0$ by taking the square root of both sides.
$4-5x = 0$
$5x = 4$
$x = \frac{4}{5}$
So we have another critical point at $x = \frac{4}{5}$ .

Solve $2(4-5x) - 15x = 0$ .
$8 - 10x - 15x = 0$
$-25x + 8 = 0$
$25x = 8$
$x = \frac{8}{25}$
So we have another critical point at $x = \frac{8}{25}$ .

Now we have all the critical points: $x = 0$ , $x = \frac{4}{5}$ , and $x = \frac{8}{25}$ .

To determine whether these critical points are relative maxima or minima, we can use the first derivative test or the second derivative test. For simplicity, we will use the first derivative test by checking the sign of $f'(x)$ around each critical point.

Create a sign chart for $f'(x)$ around the critical points. This involves picking test points between and around the critical points and substituting them into $f'(x)$ to check the sign.

Choose test points $x = -1$ , $x = \frac{1}{4}$ , $x = \frac{1}{2}$ , and $x = 1$ to check the sign of $f'(x)$ around the critical points.

From the sign chart, we can see that $f'(x)$ changes from positive to negative at $x = \frac{8}{25}$ , indicating a relative maximum at this point.

$f'(x)$ changes from negative to positive at $x = 0$ , indicating a relative minimum at this point.

$f'(x)$ changes from positive to negative at $x = \frac{4}{5}$ , indicating a relative maximum at this point.

The estimated relative extrema are a relative minimum at $x = 0$ and relative maximums at $x = \frac{8}{25}$ and $x = \frac{4}{5}$ . The graph would show a parabolic curve opening upwards with these extrema.