Answer :
To find the absolute maximum value of the function [tex]\( y = 3x^5 - 25x^3 + 60x + 1 \)[/tex] on the interval [tex]\([0, 3]\)[/tex], follow these steps:
1. Find the Derivative:
First, determine the derivative [tex]\( y' \)[/tex] of the function to find the critical points. The derivative is:
[tex]\[
y' = \frac{d}{dx}(3x^5 - 25x^3 + 60x + 1) = 15x^4 - 75x^2 + 60
\][/tex]
2. Set the Derivative Equal to Zero:
Solve the equation [tex]\( 15x^4 - 75x^2 + 60 = 0 \)[/tex] to find the critical points:
[tex]\[
15x^4 - 75x^2 + 60 = 0
\][/tex]
Let [tex]\( u = x^2 \)[/tex], then the equation becomes:
[tex]\[
15u^2 - 75u + 60 = 0
\][/tex]
Solve this quadratic equation using the quadratic formula:
[tex]\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\( a = 15 \)[/tex], [tex]\( b = -75 \)[/tex], and [tex]\( c = 60 \)[/tex]. Thus, you will find [tex]\( u_1 \)[/tex] and [tex]\( u_2 \)[/tex].
3. Convert Back to x:
When you have the solutions for [tex]\( u \)[/tex], convert back to [tex]\( x \)[/tex] by solving [tex]\( x = \pm \sqrt{u} \)[/tex].
4. Evaluate at Critical Points and Endpoints:
Examine these points within the interval [tex]\( [0, 3] \)[/tex] and also evaluate the function at the endpoints [tex]\( x=0 \)[/tex] and [tex]\( x=3 \)[/tex].
5. Find the Maximum Value:
Calculate the function [tex]\( y \)[/tex] at each of these points and determine which one gives the highest value.
In this problem, after evaluating the function at all critical points and endpoints within the given interval, the highest value you get is:
[tex]\[ 235 \][/tex]
Therefore, the absolute maximum value of the function on the interval [tex]\([0, 3]\)[/tex] is [tex]\( \boxed{235} \)[/tex].
1. Find the Derivative:
First, determine the derivative [tex]\( y' \)[/tex] of the function to find the critical points. The derivative is:
[tex]\[
y' = \frac{d}{dx}(3x^5 - 25x^3 + 60x + 1) = 15x^4 - 75x^2 + 60
\][/tex]
2. Set the Derivative Equal to Zero:
Solve the equation [tex]\( 15x^4 - 75x^2 + 60 = 0 \)[/tex] to find the critical points:
[tex]\[
15x^4 - 75x^2 + 60 = 0
\][/tex]
Let [tex]\( u = x^2 \)[/tex], then the equation becomes:
[tex]\[
15u^2 - 75u + 60 = 0
\][/tex]
Solve this quadratic equation using the quadratic formula:
[tex]\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\( a = 15 \)[/tex], [tex]\( b = -75 \)[/tex], and [tex]\( c = 60 \)[/tex]. Thus, you will find [tex]\( u_1 \)[/tex] and [tex]\( u_2 \)[/tex].
3. Convert Back to x:
When you have the solutions for [tex]\( u \)[/tex], convert back to [tex]\( x \)[/tex] by solving [tex]\( x = \pm \sqrt{u} \)[/tex].
4. Evaluate at Critical Points and Endpoints:
Examine these points within the interval [tex]\( [0, 3] \)[/tex] and also evaluate the function at the endpoints [tex]\( x=0 \)[/tex] and [tex]\( x=3 \)[/tex].
5. Find the Maximum Value:
Calculate the function [tex]\( y \)[/tex] at each of these points and determine which one gives the highest value.
In this problem, after evaluating the function at all critical points and endpoints within the given interval, the highest value you get is:
[tex]\[ 235 \][/tex]
Therefore, the absolute maximum value of the function on the interval [tex]\([0, 3]\)[/tex] is [tex]\( \boxed{235} \)[/tex].
