Question

Find the local maximum and minimum values of the function $f(x) = {x^3} + 4{x^2} - 3x + 1$

Answer 1

Hint:
At first we need to differentiate the given function and find the value of x for which ${f^{'}}(x) = 0$ and we split the number line into intervals based on the values of x and we need to select a number from each of the interval and substitute in ${f^{'}}(x)$ and based on the signs of the values we can find whether the function is increasing or decreasing and using that we can find the minimum and maximum values.

Complete step by step solution:
We are given a function $f(x) = {x^3} + 4{x^2} - 3x + 1$
We are asked to find the local minimum and local maximum values of the function
At first we need to differentiate the given function
$ \Rightarrow {f^{'}}(x) = 3{x^2} + 8x - 3$
Now we need to find the values of x for which ${f^{'}}(x) = 0$
Hence we need to equate ${f^{'}}(x) = 0$
$
   \Rightarrow 3{x^2} + 8x - 3 = 0 \\
   \Rightarrow 3{x^2} + 9x - x - 3 = 0 \\
   \Rightarrow 3x\left( {x + 3} \right) - 1\left( {x + 3} \right) = 0 \\
   \Rightarrow \left( {x + 3} \right)\left( {3x - 1} \right) = 0 \\
   \Rightarrow x = - 3,\dfrac{1}{3} \\
 $
Hence we get the values of x
Now marking the values on the number line
Now we need to take a value in the interval $\left( { - \infty , - 3} \right)$ and substituting it in $\left( {x + 3} \right)\left( {3x - 1} \right)$
So let the number be -4
Substituting we get
$ \Rightarrow \left( { - 4 + 3} \right)\left( {3\left( { - 4} \right) - 1} \right) = \left( { - 1} \right)\left( { - 13} \right) = 13$
Hence we get a positive number
Hence the function is positive in that interval
Now we need to take a value in the interval $\left( { - 3,\dfrac{1}{3}} \right)$ and substituting it in $\left( {x + 3} \right)\left( {3x - 1} \right)$
So let the number be 0
Substituting we get
$ \Rightarrow \left( {0 + 3} \right)\left( {3\left( 0 \right) - 1} \right) = \left( 3 \right)\left( { - 1} \right) = - 1$
Hence we get a negative number
Hence the function is negative in that interval
Now we need to take a value in the interval $\left( {\dfrac{1}{3},\infty } \right)$ and substituting it in $\left( {x + 3} \right)\left( {3x - 1} \right)$
So let the number be 1
Substituting we get
$ \Rightarrow \left( {1 + 3} \right)\left( {3\left( 1 \right) - 1} \right) = \left( 4 \right)\left( 2 \right) = 8$
Hence we get a positive number
Hence the function is positive in that interval
Now considering the intervals $\left( { - \infty , - 3} \right)$and $\left( { - 3,\dfrac{1}{3}} \right)$
We can see that the function changes from positive to negative
hence there is a maximum at x = -3
Now considering the intervals $\left( { - 3,\dfrac{1}{3}} \right)$and $\left( {\dfrac{1}{3},\infty } \right)$
We can see that the function changes from negative to positive
hence there is a minimum at x = $\dfrac{1}{3}$
So to find the values we need to substitute that points in f(x)
Therefore the local maximum value of x = -3
$
   \Rightarrow f( - 3) = {\left( { - 3} \right)^3} + 4{\left( { - 3} \right)^2} - 3\left( { - 3} \right) + 1 \\
   \Rightarrow f( - 3) = - 27 + 4\left( 9 \right) + 9 + 1 \\
   \Rightarrow f( - 3) = - 27 + 36 + 10 = 46 - 27 = 19 \\
 $
Hence the local maximum value of the function is 19
Therefore the local minimum value of x = $\dfrac{1}{3}$
$
   \Rightarrow f\left( {\dfrac{1}{3}} \right) = {\left( {\dfrac{1}{3}} \right)^3} + 4{\left( {\dfrac{1}{3}} \right)^2} - 3\left( {\dfrac{1}{3}} \right) + 1 \\
   \Rightarrow f\left( {\dfrac{1}{3}} \right) = \left( {\dfrac{1}{{27}}} \right) + 4\left( {\dfrac{1}{9}} \right) - 1 + 1 \\
   \Rightarrow f\left( {\dfrac{1}{3}} \right) = \dfrac{1}{{27}} + \dfrac{4}{9} \\
   \Rightarrow f\left( {\dfrac{1}{3}} \right) = \dfrac{{1 + 12}}{{27}} = \dfrac{{13}}{{27}} \\
 $

Hence the local minimum value of the function is $\dfrac{{13}}{{27}}$

Note:
1) We say that a function f(x) has a relative maximum value at x = a, if f(a) is greater than any value immediately preceding or following.
2) We call it a "relative" maximum because other values of the function may in fact be greater.
3) We say that a function f(x) has a relative minimum value at x = b, if f(b) is less than any value immediately preceding or following.
4) The value of the function, the value of y, at either a maximum or a minimum is called an extreme value.