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Answer :
Below is a complete step‐by‐step explanation for solving each part of the problem.
─────────────────────────────
. Average Rate of Change of
[tex]\[
f(x)=3x^4+2
\][/tex]
on the interval [tex]\([-2,3]\)[/tex]:
1. First compute the value of the function at the endpoints.
• When [tex]\( x=-2 \)[/tex]:
[tex]\[
f(-2)=3(-2)^4+2=3\cdot16+2=48+2=50.
\][/tex]
• When [tex]\( x=3 \)[/tex]:
[tex]\[
f(3)=3(3)^4+2=3\cdot81+2=243+2=245.
\][/tex]
2. The average rate of change over [tex]\([-2,3]\)[/tex] is given by:
[tex]\[
\text{Average Rate} = \frac{f(3)-f(-2)}{3-(-2)}=\frac{245-50}{5}=\frac{195}{5}=39.
\][/tex]
─────────────────────────────
. Instantaneous Rate of Change of
[tex]\[
f(x)=2x^3-3
\][/tex]
at [tex]\(x=2\)[/tex]:
1. The derivative [tex]\(f'(x)\)[/tex] is:
[tex]\[
f'(x)=\frac{d}{dx}\left(2x^3-3\right)=6x^2.
\][/tex]
2. Evaluating at [tex]\(x=2\)[/tex]:
[tex]\[
f'(2)=6(2)^2=6\cdot4=24.
\][/tex]
─────────────────────────────
. Derivative of
[tex]\[
y=(x^3+3)^2
\][/tex]
1. Let
[tex]\[
u(x)=x^3+3.
\][/tex]
Then [tex]\(y=u^2\)[/tex].
2. Using the chain rule:
[tex]\[
\frac{dy}{dx}=2u\cdot\frac{du}{dx}.
\][/tex]
3. Compute [tex]\(u'(x)\)[/tex]:
[tex]\[
u'(x)=\frac{d}{dx}\left(x^3+3\right)=3x^2.
\][/tex]
4. Substitute back:
[tex]\[
\frac{dy}{dx}=2(x^3+3)\cdot 3x^2=6x^2(x^3+3)=6x^5+18x^2.
\][/tex]
─────────────────────────────
. Instantaneous Rate of Change of
[tex]\[
f(x)=x^3+3x
\][/tex]
at [tex]\(x=2\)[/tex]:
1. Compute the derivative:
[tex]\[
f'(x)=\frac{d}{dx}\left(x^3+3x\right)=3x^2+3.
\][/tex]
2. Evaluate at [tex]\(x=2\)[/tex]:
[tex]\[
f'(2)=3(2)^2+3=3\cdot4+3=12+3=15.
\][/tex]
─────────────────────────────
. Instantaneous Rate of Change of
[tex]\[
f(x)=6x^2-3
\][/tex]
at [tex]\(x=2\)[/tex]:
1. The derivative is:
[tex]\[
f'(x)=\frac{d}{dx}\left(6x^2-3\right)=12x.
\][/tex]
2. Evaluating at [tex]\(x=2\)[/tex]:
[tex]\[
f'(2)=12\cdot2=24.
\][/tex]
─────────────────────────────
. Derivative of
[tex]\[
F(x)=f(x^2+4x) \cdot g(3x+1)
\][/tex]
at [tex]\(x=1\)[/tex], given that:
[tex]\[
f(5)=-2,\quad f'(5)=2,\quad g(4)=3,\quad g'(4)=5.
\][/tex]
1. Define two functions:
– [tex]\(u(x)=x^2+4x\)[/tex] so that [tex]\(f\)[/tex] is evaluated at [tex]\(u(x)\)[/tex].
– [tex]\(v(x)=3x+1\)[/tex] so that [tex]\(g\)[/tex] is evaluated at [tex]\(v(x)\)[/tex].
2. At [tex]\(x=1\)[/tex]:
• [tex]\(u(1)=1^2+4(1)=1+4=5\)[/tex].
• [tex]\(v(1)=3(1)+1=3+1=4\)[/tex].
3. Compute the derivatives:
• For [tex]\(u(x)\)[/tex]:
[tex]\[
u'(x)=2x+4\quad\text{thus}\quad u'(1)=2\cdot1+4=6.
\][/tex]
• For [tex]\(v(x)\)[/tex]:
[tex]\[
v'(x)=3\quad\text{(a constant)}.
\][/tex]
4. Using the product rule and chain rule, the derivative [tex]\(F'(x)\)[/tex] is:
[tex]\[
F'(x)=f'(u(x))\cdot u'(x)\cdot g(v(x)) + f(u(x))\cdot g'(v(x))\cdot v'(x).
\][/tex]
5. Evaluating at [tex]\(x=1\)[/tex]:
[tex]\[
F'(1)=f'(5)\cdot u'(1)\cdot g(4) + f(5)\cdot g'(4)\cdot v'(1).
\][/tex]
Substitute the given values:
[tex]\[
F'(1)=2\cdot6\cdot3+(-2)\cdot5\cdot3=36-30=6.
\][/tex]
─────────────────────────────
Summary of Answers:
1. The average rate of change is [tex]\(\boxed{39}\)[/tex].
2. The instantaneous rate of change for [tex]\(f(x)=2x^3-3\)[/tex] at [tex]\(x=2\)[/tex] is [tex]\(\boxed{24}\)[/tex].
3. The derivative of [tex]\(y=(x^3+3)^2\)[/tex] is [tex]\(\boxed{6x^5+18x^2}\)[/tex].
4. The instantaneous rate of change for [tex]\(f(x)=x^3+3x\)[/tex] at [tex]\(x=2\)[/tex] is [tex]\(\boxed{15}\)[/tex].
5. The instantaneous rate of change for [tex]\(f(x)=6x^2-3\)[/tex] at [tex]\(x=2\)[/tex] is [tex]\(\boxed{24}\)[/tex].
6. The derivative of [tex]\(F(x)=f(x^2+4x)g(3x+1)\)[/tex] at [tex]\(x=1\)[/tex] is [tex]\(\boxed{6}\)[/tex].
─────────────────────────────
. Average Rate of Change of
[tex]\[
f(x)=3x^4+2
\][/tex]
on the interval [tex]\([-2,3]\)[/tex]:
1. First compute the value of the function at the endpoints.
• When [tex]\( x=-2 \)[/tex]:
[tex]\[
f(-2)=3(-2)^4+2=3\cdot16+2=48+2=50.
\][/tex]
• When [tex]\( x=3 \)[/tex]:
[tex]\[
f(3)=3(3)^4+2=3\cdot81+2=243+2=245.
\][/tex]
2. The average rate of change over [tex]\([-2,3]\)[/tex] is given by:
[tex]\[
\text{Average Rate} = \frac{f(3)-f(-2)}{3-(-2)}=\frac{245-50}{5}=\frac{195}{5}=39.
\][/tex]
─────────────────────────────
. Instantaneous Rate of Change of
[tex]\[
f(x)=2x^3-3
\][/tex]
at [tex]\(x=2\)[/tex]:
1. The derivative [tex]\(f'(x)\)[/tex] is:
[tex]\[
f'(x)=\frac{d}{dx}\left(2x^3-3\right)=6x^2.
\][/tex]
2. Evaluating at [tex]\(x=2\)[/tex]:
[tex]\[
f'(2)=6(2)^2=6\cdot4=24.
\][/tex]
─────────────────────────────
. Derivative of
[tex]\[
y=(x^3+3)^2
\][/tex]
1. Let
[tex]\[
u(x)=x^3+3.
\][/tex]
Then [tex]\(y=u^2\)[/tex].
2. Using the chain rule:
[tex]\[
\frac{dy}{dx}=2u\cdot\frac{du}{dx}.
\][/tex]
3. Compute [tex]\(u'(x)\)[/tex]:
[tex]\[
u'(x)=\frac{d}{dx}\left(x^3+3\right)=3x^2.
\][/tex]
4. Substitute back:
[tex]\[
\frac{dy}{dx}=2(x^3+3)\cdot 3x^2=6x^2(x^3+3)=6x^5+18x^2.
\][/tex]
─────────────────────────────
. Instantaneous Rate of Change of
[tex]\[
f(x)=x^3+3x
\][/tex]
at [tex]\(x=2\)[/tex]:
1. Compute the derivative:
[tex]\[
f'(x)=\frac{d}{dx}\left(x^3+3x\right)=3x^2+3.
\][/tex]
2. Evaluate at [tex]\(x=2\)[/tex]:
[tex]\[
f'(2)=3(2)^2+3=3\cdot4+3=12+3=15.
\][/tex]
─────────────────────────────
. Instantaneous Rate of Change of
[tex]\[
f(x)=6x^2-3
\][/tex]
at [tex]\(x=2\)[/tex]:
1. The derivative is:
[tex]\[
f'(x)=\frac{d}{dx}\left(6x^2-3\right)=12x.
\][/tex]
2. Evaluating at [tex]\(x=2\)[/tex]:
[tex]\[
f'(2)=12\cdot2=24.
\][/tex]
─────────────────────────────
. Derivative of
[tex]\[
F(x)=f(x^2+4x) \cdot g(3x+1)
\][/tex]
at [tex]\(x=1\)[/tex], given that:
[tex]\[
f(5)=-2,\quad f'(5)=2,\quad g(4)=3,\quad g'(4)=5.
\][/tex]
1. Define two functions:
– [tex]\(u(x)=x^2+4x\)[/tex] so that [tex]\(f\)[/tex] is evaluated at [tex]\(u(x)\)[/tex].
– [tex]\(v(x)=3x+1\)[/tex] so that [tex]\(g\)[/tex] is evaluated at [tex]\(v(x)\)[/tex].
2. At [tex]\(x=1\)[/tex]:
• [tex]\(u(1)=1^2+4(1)=1+4=5\)[/tex].
• [tex]\(v(1)=3(1)+1=3+1=4\)[/tex].
3. Compute the derivatives:
• For [tex]\(u(x)\)[/tex]:
[tex]\[
u'(x)=2x+4\quad\text{thus}\quad u'(1)=2\cdot1+4=6.
\][/tex]
• For [tex]\(v(x)\)[/tex]:
[tex]\[
v'(x)=3\quad\text{(a constant)}.
\][/tex]
4. Using the product rule and chain rule, the derivative [tex]\(F'(x)\)[/tex] is:
[tex]\[
F'(x)=f'(u(x))\cdot u'(x)\cdot g(v(x)) + f(u(x))\cdot g'(v(x))\cdot v'(x).
\][/tex]
5. Evaluating at [tex]\(x=1\)[/tex]:
[tex]\[
F'(1)=f'(5)\cdot u'(1)\cdot g(4) + f(5)\cdot g'(4)\cdot v'(1).
\][/tex]
Substitute the given values:
[tex]\[
F'(1)=2\cdot6\cdot3+(-2)\cdot5\cdot3=36-30=6.
\][/tex]
─────────────────────────────
Summary of Answers:
1. The average rate of change is [tex]\(\boxed{39}\)[/tex].
2. The instantaneous rate of change for [tex]\(f(x)=2x^3-3\)[/tex] at [tex]\(x=2\)[/tex] is [tex]\(\boxed{24}\)[/tex].
3. The derivative of [tex]\(y=(x^3+3)^2\)[/tex] is [tex]\(\boxed{6x^5+18x^2}\)[/tex].
4. The instantaneous rate of change for [tex]\(f(x)=x^3+3x\)[/tex] at [tex]\(x=2\)[/tex] is [tex]\(\boxed{15}\)[/tex].
5. The instantaneous rate of change for [tex]\(f(x)=6x^2-3\)[/tex] at [tex]\(x=2\)[/tex] is [tex]\(\boxed{24}\)[/tex].
6. The derivative of [tex]\(F(x)=f(x^2+4x)g(3x+1)\)[/tex] at [tex]\(x=1\)[/tex] is [tex]\(\boxed{6}\)[/tex].
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