Math / Calculus

Question14. Use the table below to find the following values. \begin{tabular}{|c|c|c|c|c|} \hline $x$ & $f(x)$ & $g(x)$ & $f^{\prime}(x)$ & $g^{\prime}(x)$ \\ \hline \hline-1 & -5 & 1 & 3 & 0 \\ \hline 0 & -2 & 0 & 1 & 1 \\ \hline 1 & 0 & -3 & 0 & 0.5 \\ \hline 2 & 5 & -1 & 5 & 2 \\ \hline \end{tabular} (a) Find $h^{\prime}(2)$ whenh $(x)=-3 g(x)$ (2 pts) (b) Find $h^{\prime}(-1)$ when $h(x)=f(-1-3 x)-e^{3 x}$ (2 pts) (c) Find $h^{\prime}(2)$ when $h(x)=x^{3} g(x)(2 \mathrm{pts})$ (d) Find $h^{\prime}(0)$ whenh $(x)=g\left(-\frac{1}{2} x\right) \cdot f(x)(2 \mathrm{pts})$ (e) Find $h^{\prime}(1)$ when $h(x)=\frac{g(x)}{\left(x^{2}-4\right)}(2 \mathrm{pts})$ (f) Find $\left(f^{-1}\right)^{\prime}(-5)$ (2 pts)

Studdy Solution

STEP 1

What is this asking? We're finding the derivatives of different functions $h(x)$ at specific points using the table of values for $f(x)$ , $g(x)$ , and their derivatives. Watch out! Be careful with applying the chain rule, product rule, and quotient rule correctly.
Also, double-check the table values for accuracy!

STEP 2

1. Calculate $h^{\prime}(2)$ for $h(x) = -3g(x)$
2. Calculate $h^{\prime}(-1)$ for $h(x) = f(-1-3x) - e^{3x}$
3. Calculate $h^{\prime}(2)$ for $h(x) = x^3 g(x)$
4. Calculate $h^{\prime}(0)$ for $h(x) = g\left(-\frac{1}{2}x\right) \cdot f(x)$
5. Calculate $h^{\prime}(1)$ for $h(x) = \frac{g(x)}{x^2 - 4}$
6. Calculate $\left(f^{-1}\right)^{\prime}(-5)$

STEP 3

First, let's differentiate $h(x) = -3g(x)$ .
The derivative is:
$h^{\prime}(x) = -3g^{\prime}(x)$

STEP 4

Now, substitute $x = 2$ into the derivative:
$h^{\prime}(2) = -3g^{\prime}(2)$

STEP 5

From the table, $g^{\prime}(2) = 2$ .
So, calculate:
$h^{\prime}(2) = -3 \cdot 2 = -6$

STEP 6

Apply the chain rule to differentiate $f(-1-3x)$ :
$\frac{d}{dx} f(-1-3x) = f^{\prime}(-1-3x) \cdot (-3)$

STEP 7

Differentiate $e^{3x}$ :
$\frac{d}{dx} e^{3x} = 3e^{3x}$

STEP 8

Combine the derivatives:
$h^{\prime}(x) = f^{\prime}(-1-3x) \cdot (-3) - 3e^{3x}$

STEP 9

Substitute $x = -1$ :
$h^{\prime}(-1) = f^{\prime}(-1-3(-1)) \cdot (-3) - 3e^{3(-1)}$

STEP 10

Simplify the expression:
$h^{\prime}(-1) = f^{\prime}(2) \cdot (-3) - 3e^{-3}$

STEP 11

From the table, $f^{\prime}(2) = 5$ .
So, calculate:
$h^{\prime}(-1) = 5 \cdot (-3) - 3e^{-3} = -15 - 3e^{-3}$

STEP 12

Use the product rule:
$h^{\prime}(x) = 3x^2 g(x) + x^3 g^{\prime}(x)$

STEP 13

Substitute $x = 2$ :
$h^{\prime}(2) = 3(2)^2 g(2) + (2)^3 g^{\prime}(2)$

STEP 14

From the table, $g(2) = -1$ and $g^{\prime}(2) = 2$ .
So, calculate:
$h^{\prime}(2) = 3 \cdot 4 \cdot (-1) + 8 \cdot 2 = -12 + 16 = 4$

STEP 15

Use the product rule and chain rule:
$h^{\prime}(x) = g^{\prime}\left(-\frac{1}{2}x\right) \cdot \left(-\frac{1}{2}\right) \cdot f(x) + g\left(-\frac{1}{2}x\right) \cdot f^{\prime}(x)$

STEP 16

Substitute $x = 0$ :
$h^{\prime}(0) = g^{\prime}(0) \cdot \left(-\frac{1}{2}\right) \cdot f(0) + g(0) \cdot f^{\prime}(0)$

STEP 17

From the table, $g(0) = 0$ , $g^{\prime}(0) = 1$ , $f(0) = -2$ , and $f^{\prime}(0) = 1$ .
So, calculate:
$h^{\prime}(0) = 1 \cdot \left(-\frac{1}{2}\right) \cdot (-2) + 0 \cdot 1 = 1$

STEP 18

Use the quotient rule:
$h^{\prime}(x) = \frac{(x^2 - 4)g^{\prime}(x) - g(x) \cdot 2x}{(x^2 - 4)^2}$

STEP 19

Substitute $x = 1$ :
$h^{\prime}(1) = \frac{(1^2 - 4)g^{\prime}(1) - g(1) \cdot 2 \cdot 1}{(1^2 - 4)^2}$

STEP 20

From the table, $g(1) = -3$ and $g^{\prime}(1) = 0.5$ .
So, calculate:
$h^{\prime}(1) = \frac{(-3) \cdot 0.5 - (-3) \cdot 2}{(-3)^2} = \frac{-1.5 + 6}{9} = \frac{4.5}{9} = \frac{1}{2}$

STEP 21

Use the inverse function derivative formula:
$\left(f^{-1}\right)^{\prime}(y) = \frac{1}{f^{\prime}(x)} \quad \text{where} \quad f(x) = y$

STEP 22

From the table, $f(-1) = -5$ .
So, $x = -1$ .

STEP 23

From the table, $f^{\prime}(-1) = 3$ .
So, calculate:
$\left(f^{-1}\right)^{\prime}(-5) = \frac{1}{3}$

STEP 24

(a) $h^{\prime}(2) = -6$
(b) $h^{\prime}(-1) = -15 - 3e^{-3}$
(c) $h^{\prime}(2) = 4$
(d) $h^{\prime}(0) = 1$
(e) $h^{\prime}(1) = \frac{1}{2}$
(f) $\left(f^{-1}\right)^{\prime}(-5) = \frac{1}{3}$

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Studdy Solution

STEP 1

STEP 2

STEP 3

First, let's **differentiate** h(x)=−3g(x) h(x) = -3g(x) h(x)=−3g(x).The derivative is:h′(x)=−3g′(x)h^{\prime}(x) = -3g^{\prime}(x)h′(x)=−3g′(x)

STEP 4

Now, **substitute** x=2 x = 2 x=2 into the derivative:h′(2)=−3g′(2)h^{\prime}(2) = -3g^{\prime}(2)h′(2)=−3g′(2)

STEP 5

From the table, g′(2)=2 g^{\prime}(2) = 2 g′(2)=2.So, **calculate**:h′(2)=−3⋅2=−6h^{\prime}(2) = -3 \cdot 2 = -6h′(2)=−3⋅2=−6

STEP 6

Apply the **chain rule** to differentiate f(−1−3x) f(-1-3x) f(−1−3x):ddxf(−1−3x)=f′(−1−3x)⋅(−3)\frac{d}{dx} f(-1-3x) = f^{\prime}(-1-3x) \cdot (-3)dxd​f(−1−3x)=f′(−1−3x)⋅(−3)

STEP 7

Differentiate e3x e^{3x} e3x:ddxe3x=3e3x\frac{d}{dx} e^{3x} = 3e^{3x}dxd​e3x=3e3x

STEP 8

Combine the derivatives:h′(x)=f′(−1−3x)⋅(−3)−3e3xh^{\prime}(x) = f^{\prime}(-1-3x) \cdot (-3) - 3e^{3x}h′(x)=f′(−1−3x)⋅(−3)−3e3x

STEP 9

**Substitute** x=−1 x = -1 x=−1:h′(−1)=f′(−1−3(−1))⋅(−3)−3e3(−1)h^{\prime}(-1) = f^{\prime}(-1-3(-1)) \cdot (-3) - 3e^{3(-1)}h′(−1)=f′(−1−3(−1))⋅(−3)−3e3(−1)

STEP 10

Simplify the expression:h′(−1)=f′(2)⋅(−3)−3e−3h^{\prime}(-1) = f^{\prime}(2) \cdot (-3) - 3e^{-3}h′(−1)=f′(2)⋅(−3)−3e−3

STEP 11

From the table, f′(2)=5 f^{\prime}(2) = 5 f′(2)=5.So, **calculate**:h′(−1)=5⋅(−3)−3e−3=−15−3e−3h^{\prime}(-1) = 5 \cdot (-3) - 3e^{-3} = -15 - 3e^{-3}h′(−1)=5⋅(−3)−3e−3=−15−3e−3

STEP 12

Use the **product rule**:h′(x)=3x2g(x)+x3g′(x)h^{\prime}(x) = 3x^2 g(x) + x^3 g^{\prime}(x)h′(x)=3x2g(x)+x3g′(x)

STEP 13

**Substitute** x=2 x = 2 x=2:h′(2)=3(2)2g(2)+(2)3g′(2)h^{\prime}(2) = 3(2)^2 g(2) + (2)^3 g^{\prime}(2)h′(2)=3(2)2g(2)+(2)3g′(2)

STEP 14

From the table, g(2)=−1 g(2) = -1 g(2)=−1 and g′(2)=2 g^{\prime}(2) = 2 g′(2)=2.So, **calculate**:h′(2)=3⋅4⋅(−1)+8⋅2=−12+16=4h^{\prime}(2) = 3 \cdot 4 \cdot (-1) + 8 \cdot 2 = -12 + 16 = 4h′(2)=3⋅4⋅(−1)+8⋅2=−12+16=4

STEP 15

STEP 16

**Substitute** x=0 x = 0 x=0:h′(0)=g′(0)⋅(−12)⋅f(0)+g(0)⋅f′(0)h^{\prime}(0) = g^{\prime}(0) \cdot \left(-\frac{1}{2}\right) \cdot f(0) + g(0) \cdot f^{\prime}(0)h′(0)=g′(0)⋅(−21​)⋅f(0)+g(0)⋅f′(0)

STEP 17

STEP 18

Use the **quotient rule**:h′(x)=(x2−4)g′(x)−g(x)⋅2x(x2−4)2h^{\prime}(x) = \frac{(x^2 - 4)g^{\prime}(x) - g(x) \cdot 2x}{(x^2 - 4)^2}h′(x)=(x2−4)2(x2−4)g′(x)−g(x)⋅2x​

STEP 19

**Substitute** x=1 x = 1 x=1:h′(1)=(12−4)g′(1)−g(1)⋅2⋅1(12−4)2h^{\prime}(1) = \frac{(1^2 - 4)g^{\prime}(1) - g(1) \cdot 2 \cdot 1}{(1^2 - 4)^2}h′(1)=(12−4)2(12−4)g′(1)−g(1)⋅2⋅1​

STEP 20

STEP 21

Use the **inverse function derivative formula**:(f−1)′(y)=1f′(x)wheref(x)=y\left(f^{-1}\right)^{\prime}(y) = \frac{1}{f^{\prime}(x)} \quad \text{where} \quad f(x) = y(f−1)′(y)=f′(x)1​wheref(x)=y

STEP 22

From the table, f(−1)=−5 f(-1) = -5 f(−1)=−5.So, x=−1 x = -1 x=−1.

STEP 23

From the table, f′(−1)=3 f^{\prime}(-1) = 3 f′(−1)=3.So, **calculate**:(f−1)′(−5)=13\left(f^{-1}\right)^{\prime}(-5) = \frac{1}{3}(f−1)′(−5)=31​

STEP 24

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First, let's differentiate $h(x) = -3g(x)$ .
The derivative is:
$h^{\prime}(x) = -3g^{\prime}(x)$

Now, substitute $x = 2$ into the derivative:
$h^{\prime}(2) = -3g^{\prime}(2)$

From the table, $g^{\prime}(2) = 2$ .
So, calculate:
$h^{\prime}(2) = -3 \cdot 2 = -6$

Apply the chain rule to differentiate $f(-1-3x)$ :
$\frac{d}{dx} f(-1-3x) = f^{\prime}(-1-3x) \cdot (-3)$

Differentiate $e^{3x}$ :
$\frac{d}{dx} e^{3x} = 3e^{3x}$

Combine the derivatives:
$h^{\prime}(x) = f^{\prime}(-1-3x) \cdot (-3) - 3e^{3x}$

Substitute $x = -1$ :
$h^{\prime}(-1) = f^{\prime}(-1-3(-1)) \cdot (-3) - 3e^{3(-1)}$

Simplify the expression:
$h^{\prime}(-1) = f^{\prime}(2) \cdot (-3) - 3e^{-3}$

From the table, $f^{\prime}(2) = 5$ .
So, calculate:
$h^{\prime}(-1) = 5 \cdot (-3) - 3e^{-3} = -15 - 3e^{-3}$

Use the product rule:
$h^{\prime}(x) = 3x^2 g(x) + x^3 g^{\prime}(x)$

Substitute $x = 2$ :
$h^{\prime}(2) = 3(2)^2 g(2) + (2)^3 g^{\prime}(2)$

From the table, $g(2) = -1$ and $g^{\prime}(2) = 2$ .
So, calculate:
$h^{\prime}(2) = 3 \cdot 4 \cdot (-1) + 8 \cdot 2 = -12 + 16 = 4$

Substitute $x = 0$ :
$h^{\prime}(0) = g^{\prime}(0) \cdot \left(-\frac{1}{2}\right) \cdot f(0) + g(0) \cdot f^{\prime}(0)$

Use the quotient rule:
$h^{\prime}(x) = \frac{(x^2 - 4)g^{\prime}(x) - g(x) \cdot 2x}{(x^2 - 4)^2}$

Substitute $x = 1$ :
$h^{\prime}(1) = \frac{(1^2 - 4)g^{\prime}(1) - g(1) \cdot 2 \cdot 1}{(1^2 - 4)^2}$

Use the inverse function derivative formula:
$\left(f^{-1}\right)^{\prime}(y) = \frac{1}{f^{\prime}(x)} \quad \text{where} \quad f(x) = y$

From the table, $f(-1) = -5$ .
So, $x = -1$ .

From the table, $f^{\prime}(-1) = 3$ .
So, calculate:
$\left(f^{-1}\right)^{\prime}(-5) = \frac{1}{3}$