finding an equation of a tangent line in exercises 37 - 42, find an equation of the line that is tangent to…

finding an equation of a tangent line in exercises 37 - 42, find an equation of the line that is tangent to the graph of f and parallel to the given line. function line 37. $f(x)=-\frac{1}{4}x^{2}$ $x + y = 0$ 38. $f(x)=2x^{2}$ $4x + y + 3 = 0$ 39. $f(x)=x^{3}$ $3x - y + 1 = 0$ 40. $f(x)=x^{3}+2$ $3x - y - 4 = 0$ 41. $f(x)=\frac{1}{sqrt{x}}$ $x + 2y - 6 = 0$ 42. $f(x)=\frac{1}{sqrt{x - 1}}$ $x + 2y + 7 = 0$

finding an equation of a tangent line in exercises 37 - 42, find an equation of the line that is tangent to the graph of f and parallel to the given line. function line 37. $f(x)=-\frac{1}{4}x^{2}$ $x + y = 0$ 38. $f(x)=2x^{2}$ $4x + y + 3 = 0$ 39. $f(x)=x^{3}$ $3x - y + 1 = 0$ 40. $f(x)=x^{3}+2$ $3x - y - 4 = 0$ 41. $f(x)=\frac{1}{sqrt{x}}$ $x + 2y - 6 = 0$ 42. $f(x)=\frac{1}{sqrt{x - 1}}$ $x + 2y + 7 = 0$

Answer

Explanation:

Step1: Rewrite the line in slope - intercept form

For a line $Ax + By+C = 0$, rewrite it as $y=-Ax/B - C/B$ to find its slope. For example, for the line $x + y=0$, we can rewrite it as $y=-x$, and its slope $m=-1$.

Step2: Find the derivative of the function

The derivative of $y = f(x)=-\frac{1}{4}x^{2}$ is $f^\prime(x)=-\frac{1}{2}x$ using the power - rule $(x^n)^\prime=nx^{n - 1}$.

Step3: Set the derivative equal to the slope of the line

Since the tangent line is parallel to the given line, their slopes are equal. Set $f^\prime(x)=-\frac{1}{2}x=-1$, then solve for $x$. We get $x = 2$.

Step4: Find the corresponding $y$ - value on the function

Substitute $x = 2$ into $f(x)=-\frac{1}{4}x^{2}$, we have $f(2)=-\frac{1}{4}\times2^{2}=-1$.

Step5: Use the point - slope form to find the equation of the tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope. Here $x_1 = 2,y_1=-1,m=-1$, so the equation is $y+1=-(x - 2)$, which simplifies to $x+y - 1=0$.

Let's solve problem 37 completely:

  1. Rewrite the line $x + y=0$ in slope - intercept form:
    • Rewrite $x + y=0$ as $y=-x$. The slope of this line is $m=-1$.
  2. Find the derivative of the function $f(x)=-\frac{1}{4}x^{2}$:
    • Using the power - rule $(x^n)^\prime=nx^{n - 1}$, we have $f^\prime(x)=-\frac{1}{2}x$.
  3. Set the derivative equal to the slope of the line:
    • Set $-\frac{1}{2}x=-1$. Solving for $x$ gives $x = 2$.
  4. Find the corresponding $y$ - value on the function:
    • Substitute $x = 2$ into $f(x)=-\frac{1}{4}x^{2}$. Then $f(2)=-\frac{1}{4}\times4=-1$.
  5. Use the point - slope form to find the equation of the tangent line:
    • The point - slope form is $y - y_1=m(x - x_1)$, with $x_1 = 2,y_1=-1,m=-1$.
    • $y+1=-(x - 2)$, which simplifies to $x + y-1=0$.

Answer:

For $f(x)=-\frac{1}{4}x^{2}$ and $x + y=0$, the equation of the tangent line is $x + y-1=0$.

(We can follow the same steps for the other problems 38 - 42. For example, for problem 38:

  1. Rewrite $4x + y+3 = 0$ as $y=-4x - 3$, slope $m=-4$.
  2. $f(x)=2x^{2}$, $f^\prime(x)=4x$.
  3. Set $4x=-4$, $x=-1$.
  4. $f(-1)=2\times(-1)^{2}=2$.
  5. Using point - slope form $y - 2=-4(x + 1)$, simplifies to $4x+y+2 = 0$.)