Answer:
a. f is increasing in the interval [tex](-4,0)[/tex] and decreasing in the intervals [tex]\left ( -\infty ,-4 \right )\,,\,\left ( 0,4 \right )[/tex]
b. local maximum value of the function is 6, -250 and Local minimum value of the function is -250
c. inflexion points are [tex]\left ( \frac{-4\sqrt{3}}{3},\frac{-1226}{9} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\frac{-1226}{9} \right )[/tex]
f is concave up in intervals [tex]\left ( -\infty ,\frac{-4\sqrt{3}}{3} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\infty \right )[/tex] and concave down in interval [tex]\left (\frac{-4\sqrt{3}}{3},\frac{4\sqrt{3}}{3} \right )[/tex]
Step-by-step explanation:
Given: [tex]f(x)=x^4-32x^2+6[/tex]
To find: interval on which f is increasing and decreasing, local minimum and maximum values of f, inflection points and interval on which f is concave up and concave down
Solution:
A function f is increasing in the interval in which [tex]f'>0[/tex] and decreasing in the interval in which [tex]f''<0[/tex]
If a function is increasing before a point and decreasing after that point, the point is said to be a point of local maxima.
If a function is decreasing before a point and increasing after that point, the point is said to be a point of local minima.
An inflection point is a point on the graph of a function at which the concavity changes. Put second derivative equal to zero as check if concavity changes at the points obtained. Such points are said to be points of inflexion.
A function f is concave up in the interval in which [tex]f''>0[/tex] and concave down in the interval in which [tex]f''<0[/tex]
a.
[tex]f(x)=x^4-32x^2+6\\f'(x)=4x^3-64x\\=4x(x^2-16)\\=4x(x+4)(x-4)[/tex]
[tex]f'(x)=0\\4x(x+4)(x-4)=0[/tex]
Observe the attached table.
So, f is increasing in the interval [tex](-4,0)[/tex] and decreasing in the intervals [tex]\left ( -\infty ,-4 \right )\,,\,\left ( 0,4 \right )[/tex]
b.
From the table,
a function f has a local maxima at [tex]x=0,4[/tex] and local minima at [tex]x=-4[/tex]
[tex]f(-4)=(-4)^4-32(-4)^2+6=256-512+6=-250\\f(0)=0^4-32(0)^2+6=6\\f(4)=4^4-32(4)^2+6=256-512+6=-250[/tex]
So, local maximum value of the function is 6, -250
Local minimum value of the function is -250
c.
[tex]f'(x)=4x^3-64x\\f''(x)=12x^2-64=4(3x^2-16)\\f''(x)=0\Rightarrow 4(3x^2-16)=0\\3x^2-16=0\\x^2=\frac{16}{3}\\x=\pm \frac{4}{\sqrt{3}}=\pm \frac{4\sqrt{3}}{3}[/tex]
See the attached table
So, f is concave up in intervals [tex]\left ( -\infty ,\frac{-4\sqrt{3}}{3} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\infty \right )[/tex]
and concave down in interval [tex]\left (\frac{-4\sqrt{3}}{3},\frac{4\sqrt{3}}{3} \right )[/tex]
Also,
[tex]f\left ( \frac{-4\sqrt{3}}{3} \right )=\left ( \frac{-4\sqrt{3}}{3} \right )^4-32\left ( \frac{-4\sqrt{3}}{3} \right )^2+6=\frac{-1226}{9}\\f\left ( \frac{4\sqrt{3}}{3} \right )=\left ( \frac{4\sqrt{3}}{3} \right )^4-32\left ( \frac{4\sqrt{3}}{3} \right )^2+6=\frac{-1226}{9}[/tex]
So, inflexion points are [tex]\left ( \frac{-4\sqrt{3}}{3},\frac{-1226}{9} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\frac{-1226}{9} \right )[/tex]
