Question

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.

Explanation:

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 \((1,-3)\).

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 \((1,-3)\).