Chapter 2 - Differentiation - 2.1 Exercises - Page 104: 57

For $f(x) = x^2$, the graph of the function and the respective tangent lines are as shown as in part (a) of the attached picture. For $g(x) = x^3$, the graph of the function and the respective tangent lines are as shown as in part (b) of the attached picture. From the graphs for part (a) and part (b), we observe that the slopes of tangent lines to the graph of a function at different values of x are not always distinct.

(a) Given $f(x) = x^2$, the derivative of this function is $f'(x)=2x$. At $x=-1$, the y-coordinate of tangency is $f(-1) = (-1)^2=1$ and the slope is given by $f'(-1)=-2$. Substituting the x and y coordinates $(-1,1)$ into our initial guess for the equation of the tangent line $ y = -2x + c$, we get $ 1 = -2 * (-1) + c $. Solving for $c$ gives us $ c = -1 $. Therefore, the equation of the tangent line is $ y = -2x - 1$. At $x=0$, the y-coordinate of tangency is $f(0) = (0)^2=0$ and the slope is given by $f'(0)=2*0=0$. Substituting the x and y coordinates $(0,0)$ into our initial guess for the equation of the tangent line $ y = 0*x + c$, we get $ 0 = 0*0 + c $. Solving for $c$ gives us $ c = 0 $. Therefore, the equation of the tangent line is $ y = 0$. At $x=1$, the y-coordinate of tangency is $f(1) = (1)^2=1$ and the slope is given by $f'(1)=2*1=2$. Substituting the x and y coordinates $(1,1)$ into our initial guess for the equation of the tangent line $ y = 2*x + c$, we get $ 1 = 2*1 + c $. Solving for $c$ gives us $ c = -1 $. Therefore, the equation of the tangent line is $ y = 2x + c$. Therefore, the graph of the function and the respective tangent lines are as shown as in part (a) of the attached picture. (b) Given $g(x) = x^3$, the derivative of this function is $g'(x)=3x^2$. At $x=-1$, the y-coordinate of tangency is $g(-1) = (-1)^3=-1$ and the slope is given by $g'(-1)=3$. Substituting the x and y coordinates $(-1,-1)$ into our initial guess for the equation of the tangent line $ y = 3x + c$, we get $ -1 = 3 * (-1) + c $. Solving for $c$ gives us $ c = 4 $. Therefore, the equation of the tangent line is $ y = 3x + 4$. At $x=0$, the y-coordinate of tangency is $g(0) = (0)^3=0$ and the slope is given by $g'(0)=0$. Substituting the x and y coordinates $(0,0)$ into our initial guess for the equation of the tangent line $ y = 0*x + c$, we get $ 0 = 0 * (0) + c $. Solving for $c$ gives us $ c = 0 $. Therefore, the equation of the tangent line is $ y = 0$. At $x=1$, the y-coordinate of tangency is $g(1) = (1)^3=1$ and the slope is given by $g'(1)=3*(1)^2=3$. Substituting the x and y coordinates $(1,1)$ into our initial guess for the equation of the tangent line $ y = 3*x + c$, we get $ 1= 3*(1) + c $. Solving for $c$ gives us $ c = -2 $. Therefore, the equation of the tangent line is $ y = 3x-2$. Therefore, the graph of the function and the respective tangent lines are as shown as in part (b) of the attached picture. From the graphs for part (a) and part (b), we observe that the slopes of tangent lines to the graph of a function at different values of x are not always distinct.