answer the following questions (6 - 10) for $f(x)=x^{3}-4x$. 6. sketch a graph of $y = f(x)$. 7. use the…

answer the following questions (6 - 10) for $f(x)=x^{3}-4x$. 6. sketch a graph of $y = f(x)$. 7. use the definition of the derivative to find $f(x)$. 8. find the slope of the curve at $x=-2$ and find the slope of the curve at $x = 3$. 9. find the equation of the tangent line for $f(x)$ at $x = 1$. show all work!! 10. sketch the graph of #9 on the same graph of the function.

answer the following questions (6 - 10) for $f(x)=x^{3}-4x$. 6. sketch a graph of $y = f(x)$. 7. use the definition of the derivative to find $f(x)$. 8. find the slope of the curve at $x=-2$ and find the slope of the curve at $x = 3$. 9. find the equation of the tangent line for $f(x)$ at $x = 1$. show all work!! 10. sketch the graph of #9 on the same graph of the function.

Answer

6. Sketch a graph of (y = f(x)=x^{3}-4x)

  • Find the roots:
    • Set (y = 0), then (x^{3}-4x=x(x^{2} - 4)=x(x - 2)(x + 2)=0). The roots are (x=-2,0,2).
    • Take the first - derivative (f'(x)=3x^{2}-4). Set (f'(x)=0), then (3x^{2}-4 = 0), (x=\pm\frac{2}{\sqrt{3}}\approx\pm1.15).
    • Take the second - derivative (f''(x)=6x). When (x =-\frac{2}{\sqrt{3}}), (f''(x)<0), so there is a local maximum at (x =-\frac{2}{\sqrt{3}}), (y=f(-\frac{2}{\sqrt{3}})=\frac{16}{3\sqrt{3}}\approx3.08). When (x=\frac{2}{\sqrt{3}}), (f''(x)>0), so there is a local minimum at (x=\frac{2}{\sqrt{3}}), (y =-\frac{16}{3\sqrt{3}}\approx - 3.08).
    • As (x\to\pm\infty), (y\to\pm\infty) since the leading term is (x^{3}).

7. Use the definition of the derivative to find (f'(x))

  • The definition of the derivative is (f'(x)=\lim_{h\to0}\frac{f(x + h)-f(x)}{h}).
    • Given (f(x)=x^{3}-4x), then (f(x + h)=(x + h)^{3}-4(x + h)=x^{3}+3x^{2}h + 3xh^{2}+h^{3}-4x-4h).
    • (f(x + h)-f(x)=x^{3}+3x^{2}h + 3xh^{2}+h^{3}-4x-4h-(x^{3}-4x)=3x^{2}h + 3xh^{2}+h^{3}-4h).
    • (\frac{f(x + h)-f(x)}{h}=\frac{3x^{2}h + 3xh^{2}+h^{3}-4h}{h}=3x^{2}+3xh + h^{2}-4).
    • Taking the limit as (h\to0), (f'(x)=\lim_{h\to0}(3x^{2}+3xh + h^{2}-4)=3x^{2}-4).

8. Find the slope of the curve at (x=-2) and (x = 3)

  • Since (f'(x)=3x^{2}-4).
    • When (x=-2), (f'(-2)=3\times(-2)^{2}-4=3\times4 - 4=8).
    • When (x = 3), (f'(3)=3\times3^{2}-4=3\times9-4=23).

9. Find the equation of the tangent line for (f(x)) at (x = 1)

  • First, find (f(1)): (f(1)=1^{3}-4\times1=-3).
  • Then, find (f'(1)): (f'(x)=3x^{2}-4), so (f'(1)=3\times1^{2}-4=-1).
  • The point - slope form of a line is (y - y_{1}=m(x - x_{1})), where ((x_{1},y_{1})=(1,-3)) and (m=-1).
  • The equation of the tangent line is (y+3=-(x - 1)), which simplifies to (y=-x - 2).

10. Sketch the graph of the tangent line (y=-x - 2) on the same graph of the function

  • The tangent line (y=-x - 2) is a straight line with a slope of (-1) and a (y) - intercept of (-2). Plot the points on the function (y=x^{3}-4x) and the tangent line (y=-x - 2) on the same coordinate axes. The tangent line touches the curve (y=x^{3}-4x) at the point ((1,-3)).