Answer :
Step-by-step explanation:
f(x) = 2x³ - 5x² + 3x - 7
1. f'(x) = 3×2x² - 2×5x + 3 = 6x² - 10x + 3
2. a critical point of f(x) is any x-value, where f'(x) = 0 or f(x) is not differentiable.
f(x) is a normal, continuous function and is everywhere differentiable.
so, we need to look for f'(x) = 0.
6x² - 10x + 3 = 0
a quadratic equation
ax² + bx + c = 0
has the generic solution
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 6
b = -10
c = 3
x = (10 ± sqrt((-10)² - 4×6×3))/(2×6) =
= (10 ± sqrt(100 - 72))/12 = (10 ± sqrt(28))/12 =
= (10 ± sqrt(4×7))/12 = (10 ± 2×sqrt(7))/12 =
= (5 ± sqrt(7))/6
x1 = (5 + sqrt(7))/6 = 1.274291885...
x2 = (5 - sqrt(7))/6 = 0.392374781...
the critical points are (local) maximums or minimums.
3. the second derivative tells us, if the maximum or minimum points are maximums or minimums.
f"(x1 or x2) > 0, it is a (local) minimum.
f"(x1 or x2) < 0, it is a (local) maximum.
f"(x) = 12x - 10
f"(x1) = 5.291502622... > 0, it is a minimum.
f"(x2) = -5.291502622... < 0, it is a maximum.
(-infinity, x2) increasing.
(x2, x1) decreasing
(x1, +infinity) increasing
4.
as found in 2. and 3.
x1 is a local minimum.
x2 is a local maximum.
Answer:
[tex]\textsf{1)}\quad f'(x)=6x^2-10x+3[/tex]
[tex]\textsf{2)}\quad \textsf{Critical points:}\quad (1.274, -7.158)\;\;\textsf{and}\;\;(0.392, -6.472)[/tex]
[tex]\textsf{3)}\quad \textsf{Increasing:}\quad (-\infty, 0.392)\;\;\textsf{and}\;\;(1.274, \infty)\\\phantom{w.w}\textsf{Decreasing:}\quad (0.392,1.274)[/tex]
[tex]\textsf{4)}\quad \textsf{Local maximum:} \quad (0.392, -6.472)\\\phantom{wdo}\textsf{Local minimum:} \quad\; (1.274, -7.158)[/tex]
Step-by-step explanation:
Question 1
To find the derivative of the function f(x) = 2x³ - 5x² + 3x - 7 with respect to x, we apply the power rule for differentiation.
[tex]\boxed{\begin{array}{c}\underline{\textsf{Power Rule for Differentiation}}\\\\\dfrac{\text{d}}{\text{d}x}\left(ax^n\right)=nax^{n-1}\\\\\textsf{where $a$ and $n$ are constants}\end{array}}[/tex]
Therefore:
[tex]f'(x)=3 \cdot 2x^{3-1}-2 \cdot 5x^{2-1}+1 \cdot 3x^{1-1}-0\\\\\\\boxed{\boxed{f'(x)=6x^2-10x+3}}[/tex]
[tex]\dotfill[/tex]
Question 2
Critical points occur where the derivative is zero or undefined. Therefore, to find the critical points of the function f(x), set f'(x) equal to zero and solve for x.
[tex]6x^2-10x+3=0[/tex]
To solve for x, use the quadratic formula:
[tex]\boxed{\begin{array}{l}\underline{\sf Quadratic\;Formula}\\\\x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\textsf{when} \;ax^2+bx+c=0 \\\end{array}}[/tex]
In this case, a = 6, b = -10 and c = 3. Therefore:
[tex]x=\dfrac{-(-10)\pm \sqrt{(-10)^2-4(6)(3)}}{2(6)} \\\\\\ x=\dfrac{10\pm \sqrt{100-72}}{12} \\\\\\ x=\dfrac{10\pm \sqrt{28}}{12} \\\\\\ x=\dfrac{10\pm \sqrt{2^2 \cdot 7}}{12} \\\\\\ x=\dfrac{10\pm \sqrt{2^2} \sqrt{7}}{12} \\\\\\ x=\dfrac{10\pm 2 \sqrt{7}}{12} \\\\\\ x=\dfrac{5\pm\sqrt{7}}{6}[/tex]
Substitute the x-values into the function to find the corresponding y-values:
[tex]f\left(\dfrac{5+\sqrt{7}}{6}\right) = 2\left(\dfrac{5+\sqrt{7}}{6}\right)^3 - 5\left(\dfrac{5+\sqrt{7}}{6}\right)^2 + 3\left(\dfrac{5+\sqrt{7}}{6}\right) - 7\\\\\\f\left(\dfrac{5+\sqrt{7}}{6}\right) = -7.15778257...[/tex]
[tex]f\left(\dfrac{5-\sqrt{7}}{6}\right) = 2\left(\dfrac{5-\sqrt{7}}{6}\right)^3 - 5\left(\dfrac{5-\sqrt{7}}{6}\right)^2 + 3\left(\dfrac{5-\sqrt{7}}{6}\right) - 7\\\\\\f\left(\dfrac{5-\sqrt{7}}{6}\right) = -6.47184705...[/tex]
Therefore, the critical points of the function rounded to three decimal places are:
[tex]\boxed{\boxed{(1.274, -7.158)\;\;\textsf{and}\;\;(0.392, -6.472)}}[/tex]
[tex]\dotfill[/tex]
Question 3
A function is increasing when f'(x) > 0.
A function is decreasing when f'(x) < 0.
Since the differentiated function f′(x) = 6x² - 10x + 3 is a quadratic polynomial with a positive leading coefficient, it forms a parabola opening upwards. This means f′(x) will be greater than zero outside the interval between the two critical points and less than zero within the interval between the critical points. Therefore:
The function f(x) is increasing on the intervals:
[tex]\left( -\infty, \dfrac{5 - \sqrt{7}}{6} \right)\;\;\textsf{and}\;\;\left( \dfrac{5 + \sqrt{7}}{6}, \infty \right)[/tex]
The function f(x) is decreasing on the interval:
[tex]\left( \dfrac{5 - \sqrt{7}}{6}, \dfrac{5 + \sqrt{7}}{6} \right)[/tex]
The intervals rounded to three decimal places are:
[tex]\boxed{\boxed{\begin{array}{l}\textsf{Increasing:}\quad (-\infty, 0.392)\;\;\textsf{and}\;\;(1.274, \infty)\\\\\textsf{Decreasing:}\quad (0.392,1.274)\end{array}}}[/tex]
[tex]\dotfill[/tex]
Question 4
Stationary points occur where f′(x) = 0.
We have already found that f′(x) = 0 when:
[tex]x=\dfrac{5+\sqrt{7}}{6}\;\;\textsf{and}\;\;x=\dfrac{5-\sqrt{7}}{6}[/tex]
Therefore, to determine if the stationary points are a minimum or a maximum, find the second derivative of the function then input the x-value of the stationary point.
- If f''(x) > 0 then its a minimum.
- If f''(x) < 0 then its a maximum.
Find the second derivative:
[tex]f''(x)=2 \cdot 6x^{2-1} - 1\cdot 10x^1-1}+0\\\\\\f''(x)=12x - 10[/tex]
Substitute the x-values when f'(x) = 0 into f''(x):
[tex]f''\left(\dfrac{5+\sqrt{7}}{6}\right)=12\left(\dfrac{5+\sqrt{7}}{6}\right)-10=2\sqrt{7} > 0 \implies \textsf{minimum}\\\\\\f''\left(\dfrac{5-\sqrt{7}}{6}\right)=12\left(\dfrac{5-\sqrt{7}}{6}\right)-10=-2\sqrt{7} < 0 \implies \textsf{maximum}[/tex]
Therefore:
[tex]\boxed{\boxed{\begin{array}{l}\textsf{Local maximum:} \quad (0.392, -6.472)\\\\\textsf{Local minimum:} \quad (1.274, -7.158)\end{array}}}[/tex]
