20
Consider $f(x)=\frac{1}{100}(3 x-5)(5 x+21)$
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 $g(x)=(2 a x-5 a)\left(x+a^{2}\right)$, 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 ：\(f'(x) = \frac{1}{100}((3x - 5)(5) + (3)(5x + 21))\)

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

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

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

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

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

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

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

Step 10 ：\(f'(x_1) + f'(x_2) = \frac{50}{3} - \frac{50}{3} = \boxed{0}\)

Step 11 ：\(g(x) = (2ax - 5a)(x + a^2)\)

Step 12 ：\(g'(x) = (2ax - 5a)(1) + (2a)(x + a^2)\)

Step 13 ：\(g'(x) = 2ax - 5a + 2ax^2 + 2a^3\)

Step 14 ：\(g'(1) = 2a(1) - 5a + 2a^2(1) + 2a^3 = 0\)

Step 15 ：\(a = \boxed{\frac{5}{9}}\)

Step 16 ：\(g'(1) = 0\), 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 \(x = 1\)