20
Consider $f(x)=\frac{1}{100}(3x-5)(5x+21)$
a.
a. Calculate $f^{\prime }(x)$.
b.
c.
b. Calculate the gradient of the tangent to the curve $y=f(x)$ at its $y$-intercept.
c. Show that the sum of the gradients of the tangents to the curve $y=f(x)$ at its $x$-intercepts is equal to zero.
d. Let $g(x)=(2ax-5a)(x+a^2)$, where $a$ is a positive real constant.
i. Calculate $g^{\prime }(x)$.
ii. For what value of $a$ does $g^{\prime }(1)=0$ ?
iii. For this value of $a$, describe the tangent to the curve $y=g(x)$ and the behaviour of the curve at the point where $x=1$

Step 1 ：$f(x)=\frac{1}{100}(3x-5)(5x+21)$

Step 2 ：$f^{\prime }(x)=\frac{1}{100}((3x-5)(5)+(3)(5x+21))$

Step 3 ：$f^{\prime }(x)=\frac{1}{100}(15x-25+15x+63)$

Step 4 ：$f^{\prime }(x)=\frac{1}{100}(30x+38)$

Step 5 ：$f(0)=\frac{1}{100}(3(0)-5)(5(0)+21)=-\frac{105}{100}$

Step 6 ：$f^{\prime }(0)=\frac{1}{100}(30(0)+38)=\frac{19}{50}$

Step 7 ：$x_1=\frac{5}{3}$, $x_2=-\frac{21}{5}$

Step 8 ：$f^{\prime }(x_1)=\frac{1}{100}(30(\frac{5}{3})+38)=\frac{50}{3}$

Step 9 ：$f^{\prime }(x_2)=\frac{1}{100}(30(-\frac{21}{5})+38)=-\frac{50}{3}$

Step 10 ：$f^{\prime }(x_1)+f^{\prime }(x_2)=\frac{50}{3}-\frac{50}{3}=\boxed{{\displaystyle 0}}$

Step 11 ：$g(x)=(2ax-5a)(x+a^2)$

Step 12 ：$g^{\prime }(x)=(2ax-5a)(1)+(2a)(x+a^2)$

Step 13 ：$g^{\prime }(x)=2ax-5a+2a{x}^2+2a^3$

Step 14 ：$g^{\prime }(1)=2a(1)-5a+2a^2(1)+2a^3=0$

Step 15 ：$a=\boxed{{\displaystyle \frac{5}{9}}}$

Step 16 ：$g^{\prime }(1)=0$, tangent is horizontal at $x=1$

Step 17 ：$g(1)=(2(\frac{5}{9})(1)-5(\frac{5}{9}))(1+(\frac{5}{9}{)}^2)$

Step 18 ：$g(1)=\frac{5}{9}(1+\frac{25}{81})$

Step 19 ：$g(1)=\frac{5}{9}(\frac{106}{81})$

Step 20 ：$g(1)=\frac{530}{729}$

Step 21 ：Tangent: $y=\frac{530}{729}$, curve has a local minimum at $x=1$