 # 3.8.1 Integration Long Questions (Question 1 & 2)

Question 1:
A curve with gradient function $5x-\frac{5}{{x}^{2}}$  has a turning point at (m, 9).
(a) Find the value of m.
(b) Determine whether the turning point is a maximum or a minimum point.
(c) Find the equation of the curve.

Solution:
(a)
$\begin{array}{l}\frac{dy}{dx}=5x-\frac{5}{{x}^{2}}\\ \text{At turning point}\left(m,9\right),\text{}\frac{dy}{dx}=0.\\ 5m-\frac{5}{{m}^{2}}=0\\ \frac{5}{{m}^{2}}=5m\\ {m}^{3}=1\\ m=1\end{array}$

(b)

(c)

Question 2:
A curve has a gradient function kx2– 7x, where k is a constant. The tangent to the curve at the point (1, 3 ) is parallel to the straight line  y + x– 4 = 0.
Find
(a) the value of k,
(b) the equation of the curve.

Solution:
(a)
y + x – 4 = 0
y = – x + 4
m = –1

f ’(x) = kx² – 7x
Given tangent to the curve at the point (1, 3) parallel to the straight line
k (1)² – 7 (1) = –1
k – 7 = –1
k = 6

(b)
$\begin{array}{l}\text{}f‘\left(x\right)=6{x}^{2}-7x\\ f\left(x\right)=\int \left(6{x}^{2}-7x\right)\text{}dx\\ f\left(x\right)=\frac{6{x}^{3}}{3}-\frac{7{x}^{2}}{2}+c\\ 3=2{\left(1\right)}^{3}-\frac{7{\left(1\right)}^{2}}{2}+c\text{at point}\left(1,3\right)\\ c=\frac{9}{2}\\ \therefore f\left(x\right)=2{x}^{3}-\frac{7{x}^{2}}{2}+\frac{9}{2}\end{array}$