Questions: Show that the curve of y=3 e^x+7 x+4 x^3 has no tangent line with slope 6. First, calculate the derivative y'. y'= Since 3 e^x > 0 and 12 x^2 > 0, we must have y > 0 0+7+0=7. So, no tangent line can have slope 6 since 6 ≠ 7. Draw a diagram to show that there are two tangent lines to the part of y=x^2 that pass through the point on the parabola. smaller x-value (x, y)=() (x, y)=() larger x-value (a) For what values of x is the function f(x)=x^2-4 differentiable? (Enter your answer using interval notation.) x=

Show that the curve of y=3 e^x+7 x+4 x^3 has no tangent line with slope 6. First, calculate the derivative y'. y'= Since 3 e^x > 0 and 12 x^2 > 0, we must have y > 0 0+7+0=7. So, no tangent line can have slope 6 since 6 ≠ 7. Draw a diagram to show that there are two tangent lines to the part of y=x^2 that pass through the point on the parabola. smaller x-value (x, y)=() (x, y)=() larger x-value (a) For what values of x is the function f(x)=x^2-4 differentiable? (Enter your answer using interval notation.) x=
Transcript text: Show that the curve of $y=3 e^{x}+7 x+4 x^{3}$ has no tangent line with slope 6 . First, calculate the derivative $y^{\prime}$. $y^{\prime}=$ $\square$ Since $3 \mathrm{e}^{x}$ ? 0 and $12 x^{2}$ $\qquad$ 0 , we must have $y$ $\qquad$ 0 $0+7+0=7$. So, no tangent line can have slope 6 since 6 $?$ 7. Draw a diagram to show that there fore two tangent lines to the part of $y=x^{2}$ that pass through the poordin parabola. smaller $x$-value \[ \begin{array}{l} (x, y)=(\square) \\ (x, y)=(\square) \end{array} \] larger $x$-value (a) For what values of $x$ is the function $f(x)=\left|x^{2}-4\right|$ differentiable? (Enter your answer using interval notation.) \[ x=\square \]
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Solution Steps

Solution Approach
Question 16

To show that the curve \( y = 3e^x + 7x + 4x^3 \) has no tangent line with slope 6, we need to find the derivative \( y' \) and check if it can ever equal 6. The derivative of \( y \) is found by differentiating each term separately.

Question 17

To find the points where the tangent lines to the parabola \( y = x^2 \) pass through a given point, we need to set up the equation of the tangent line and solve for the points of tangency.

Question 18

To determine where the function \( f(x) = |x^2 - 4| \) is differentiable, we need to check the points where the expression inside the absolute value changes sign, as these are the points where the function might not be differentiable.

Step 1: Calculate the Derivative of \( y = 3e^x + 7x + 4x^3 \)

To find the derivative \( y' \), we apply the rules of differentiation to each term in the function \( y \).

\[ y = 3e^x + 7x + 4x^3 \]

The derivative of \( 3e^x \) is \( 3e^x \).

The derivative of \( 7x \) is \( 7 \).

The derivative of \( 4x^3 \) is \( 12x^2 \).

Combining these, we get:

\[ y' = 3e^x + 7 + 12x^2 \]

Step 2: Analyze the Derivative for Slope 6

We need to determine if there is any value of \( x \) such that \( y' = 6 \).

\[ 3e^x + 7 + 12x^2 = 6 \]

Simplify the equation:

\[ 3e^x + 12x^2 + 7 = 6 \]

\[ 3e^x + 12x^2 + 1 = 0 \]

Step 3: Check for Real Solutions

We need to check if the equation \( 3e^x + 12x^2 + 1 = 0 \) has any real solutions.

Since \( 3e^x \) is always positive for all real \( x \) and \( 12x^2 \) is always non-negative, the sum \( 3e^x + 12x^2 \) is always positive. Adding 1 to a positive number will never yield zero.

Thus, there are no real solutions to the equation \( 3e^x + 12x^2 + 1 = 0 \).

Final Answer

\[ \boxed{\text{No tangent line can have slope 6.}} \]

Step 1: Identify the Points of Tangency for \( y = x^2 \)

We need to find the points on the parabola \( y = x^2 \) where the tangent lines pass through a given point. Let's assume the point is \( (a, b) \).

Step 2: Equation of the Tangent Line

The equation of the tangent line to the parabola \( y = x^2 \) at a point \( (x_1, y_1) \) is:

\[ y - y_1 = m(x - x_1) \]

where \( m \) is the slope of the tangent line. For the parabola \( y = x^2 \), the slope at \( x_1 \) is \( 2x_1 \).

So, the equation becomes:

\[ y - x_1^2 = 2x_1(x - x_1) \]

Step 3: Solve for Points of Tangency

We need to find the points \( (x_1, x_1^2) \) such that the tangent line passes through \( (a, b) \).

\[ b - x_1^2 = 2x_1(a - x_1) \]

Rearrange and solve for \( x_1 \):

\[ b - x_1^2 = 2ax_1 - 2x_1^2 \]

\[ b = 2ax_1 - x_1^2 \]

\[ x_1^2 - 2ax_1 + b = 0 \]

This is a quadratic equation in \( x_1 \). Solving for \( x_1 \):

\[ x_1 = \frac{2a \pm \sqrt{(2a)^2 - 4 \cdot 1 \cdot b}}{2 \cdot 1} \]

\[ x_1 = a \pm \sqrt{a^2 - b} \]

Final Answer

\[ \boxed{(x, y) = (a + \sqrt{a^2 - b}, (a + \sqrt{a^2 - b})^2)} \] \[ \boxed{(x, y) = (a - \sqrt{a^2 - b}, (a - \sqrt{a^2 - b})^2)} \]

Step 1: Determine Differentiability of \( f(x) = |x^2 - 4| \)

The function \( f(x) = |x^2 - 4| \) is differentiable wherever the expression inside the absolute value is not zero.

Step 2: Find Critical Points

Set the expression inside the absolute value to zero:

\[ x^2 - 4 = 0 \]

\[ x^2 = 4 \]

\[ x = \pm 2 \]

Step 3: Determine Intervals of Differentiability

The function \( f(x) \) is differentiable on the intervals where \( x^2 - 4 \neq 0 \):

\[ (-\infty, -2) \cup (-2, 2) \cup (2, \infty) \]

Final Answer

\[ \boxed{(-\infty, -2) \cup (-2, 2) \cup (2, \infty)} \]

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