Question

(a) find ( f(x) ) by the definition ( f(x)=limlimits_{h\to 0} \frac{f(x+h)-f(x)}{h} );
( f(x)= \boxed{x} ) (with a checkmark)

(b) ( f(-1)= \boxed{-1} ) (checkmark), ( f(0)= \boxed{0} ) (checkmark), ( f(2)= \boxed{2} ) (checkmark)

(c)(i) find the equation of the tangent line at the point ( (-1, f(-1)) ):
( y= \boxed{} ) (with a red x)

(c)(ii) find the equation of the tangent line at the point ( (0, f(0)) ):
( y= \boxed{} )

(c)(iii) find the equation of the tangent line at the point ( (2, f(2)) ):
( y= \boxed{} )

Explanation:

To solve the problem, we assume the function \( f(x) \) is \( \frac{1}{2}x^2 \) (since \( f'(x) = x \) from part (a)). We will find the equations of the tangent lines at the given points using the point - slope form \( y - y_1=m(x - x_1) \), where \( m = f'(x_1) \) and \( (x_1,y_1)=(x_1,f(x_1)) \)

(c)(i) Equation of the tangent line at \((- 1,f(-1))\)

1. First, find \( f(-1) \):

• Given \( f(x)=\frac{1}{2}x^{2} \), substitute \( x=-1 \) into \( f(x) \). Then \( f(-1)=\frac{1}{2}\times(-1)^{2}=\frac{1}{2} \)

1. Next, find the slope \( m \) of the tangent line at \( x = - 1 \):

• From part (a), \( f'(x)=x \), so when \( x=-1 \), \( m = f'(-1)=-1 \)

1. Now, use the point - slope form \( y - y_1=m(x - x_1) \), where \( x_1=-1,y_1 = f(-1)=\frac{1}{2} \) and \( m=-1 \)

• Substitute these values into the formula: \( y-\frac{1}{2}=-1(x + 1) \)
• Expand the right - hand side: \( y-\frac{1}{2}=-x - 1 \)
• Add \( \frac{1}{2} \) to both sides: \( y=-x - 1+\frac{1}{2}=-x-\frac{1}{2} \)

(c)(ii) Equation of the tangent line at \((0,f(0))\)

1. First, find \( f(0) \):

• Given \( f(x)=\frac{1}{2}x^{2} \), substitute \( x = 0 \) into \( f(x) \). Then \( f(0)=\frac{1}{2}\times0^{2}=0 \)

1. Next, find the slope \( m \) of the tangent line at \( x = 0 \):

• From part (a), \( f'(x)=x \), so when \( x = 0 \), \( m=f'(0)=0 \)

1. Now, use the point - slope form \( y - y_1=m(x - x_1) \), where \( x_1 = 0,y_1=f(0)=0 \) and \( m = 0 \)

• Substitute these values into the formula: \( y-0=0\times(x - 0) \)
• Simplify the equation: \( y = 0 \)

(c)(iii) Equation of the tangent line at \((2,f(2))\)

1. First, find \( f(2) \):

• Given \( f(x)=\frac{1}{2}x^{2} \), substitute \( x = 2 \) into \( f(x) \). Then \( f(2)=\frac{1}{2}\times2^{2}=2 \)

1. Next, find the slope \( m \) of the tangent line at \( x = 2 \):

• From part (a), \( f'(x)=x \), so when \( x = 2 \), \( m=f'(2)=2 \)

1. Now, use the point - slope form \( y - y_1=m(x - x_1) \), where \( x_1 = 2,y_1=f(2)=2 \) and \( m = 2 \)

• Substitute these values into the formula: \( y - 2=2(x - 2) \)
• Expand the right - hand side: \( y-2=2x-4 \)
• Add 2 to both sides: \( y=2x-4 + 2=2x-2 \)

Final Answers

(c)(i) \( y=-x-\frac{1}{2} \)

(c)(ii) \( y = 0 \)

(c)(iii) \( y=2x - 2 \)

Answer:

To solve the problem, we assume the function \( f(x) \) is \( \frac{1}{2}x^2 \) (since \( f'(x) = x \) from part (a)). We will find the equations of the tangent lines at the given points using the point - slope form \( y - y_1=m(x - x_1) \), where \( m = f'(x_1) \) and \( (x_1,y_1)=(x_1,f(x_1)) \)

(c)(i) Equation of the tangent line at \((- 1,f(-1))\)

1. First, find \( f(-1) \):

• Given \( f(x)=\frac{1}{2}x^{2} \), substitute \( x=-1 \) into \( f(x) \). Then \( f(-1)=\frac{1}{2}\times(-1)^{2}=\frac{1}{2} \)

1. Next, find the slope \( m \) of the tangent line at \( x = - 1 \):

• From part (a), \( f'(x)=x \), so when \( x=-1 \), \( m = f'(-1)=-1 \)

1. Now, use the point - slope form \( y - y_1=m(x - x_1) \), where \( x_1=-1,y_1 = f(-1)=\frac{1}{2} \) and \( m=-1 \)

• Substitute these values into the formula: \( y-\frac{1}{2}=-1(x + 1) \)
• Expand the right - hand side: \( y-\frac{1}{2}=-x - 1 \)
• Add \( \frac{1}{2} \) to both sides: \( y=-x - 1+\frac{1}{2}=-x-\frac{1}{2} \)

(c)(ii) Equation of the tangent line at \((0,f(0))\)

1. First, find \( f(0) \):

• Given \( f(x)=\frac{1}{2}x^{2} \), substitute \( x = 0 \) into \( f(x) \). Then \( f(0)=\frac{1}{2}\times0^{2}=0 \)

1. Next, find the slope \( m \) of the tangent line at \( x = 0 \):

• From part (a), \( f'(x)=x \), so when \( x = 0 \), \( m=f'(0)=0 \)

1. Now, use the point - slope form \( y - y_1=m(x - x_1) \), where \( x_1 = 0,y_1=f(0)=0 \) and \( m = 0 \)

• Substitute these values into the formula: \( y-0=0\times(x - 0) \)
• Simplify the equation: \( y = 0 \)

(c)(iii) Equation of the tangent line at \((2,f(2))\)

1. First, find \( f(2) \):

• Given \( f(x)=\frac{1}{2}x^{2} \), substitute \( x = 2 \) into \( f(x) \). Then \( f(2)=\frac{1}{2}\times2^{2}=2 \)

1. Next, find the slope \( m \) of the tangent line at \( x = 2 \):

• From part (a), \( f'(x)=x \), so when \( x = 2 \), \( m=f'(2)=2 \)

1. Now, use the point - slope form \( y - y_1=m(x - x_1) \), where \( x_1 = 2,y_1=f(2)=2 \) and \( m = 2 \)

• Substitute these values into the formula: \( y - 2=2(x - 2) \)
• Expand the right - hand side: \( y-2=2x-4 \)
• Add 2 to both sides: \( y=2x-4 + 2=2x-2 \)

Final Answers

(c)(i) \( y=-x-\frac{1}{2} \)

(c)(ii) \( y = 0 \)

(c)(iii) \( y=2x - 2 \)