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The derivative of the function
f is defined by
f^(')(x)=x^(3)-5+2sin(2x-5). Find the
x values, if any, in the interval
-0.5 < x < 2.5 where the function
f has a relative maximum. You may use a calculator and round all values to 3 decimal places.
Answer:
x=

The derivative of the function $f$ is defined by $f^{\prime}(x)=x^{3}-5+2 \sin (2 x-5)$ . Find the $x$ values, if any, in the interval $-0.5<x<2.5$ where the function $f$ has a relative maximum. You may use a calculator and round all values to $3$ decimal places. $\newline$ Answer: $x=$

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Find derivatives using logarithmic differentiation

Full solution

Q. The derivative of the function

f

is defined by

f^{\prime}(x)=x^{3}-5+2 \sin (2 x-5)

. Find the

x

values, if any, in the interval

-0.5<x<2.5

where the function

f

has a relative maximum. You may use a calculator and round all values to

3

decimal places.

\newline

Answer:

x=

Find Critical Points: To find the relative maximum of the function $f$ , we need to find the critical points of $f^{\prime}(x)$ in the given interval. Critical points occur where the derivative is zero or undefined. The derivative $f^{\prime}(x) = x^{3} - 5 + 2\sin(2x - 5)$ is defined for all $x$ , so we only need to find where it is zero.
Solve for $x$ : Set the derivative equal to zero and solve for $x$ : $x^{3} - 5 + 2\sin(2x - 5) = 0$ This equation is not easily solvable algebraically, so we will use a calculator to find the approximate values of $x$ in the interval $-0.5 < x < 2.5$ .
Graph Function: Using a calculator, we graph the function $f'(x) = x^{3} - 5 + 2\sin(2x - 5)$ and look for points where the graph crosses the x-axis within the interval $-0.5 < x < 2.5$ .
Check Sign Change: After graphing, we find that the derivative crosses the x-axis at approximately $x = 1.047$ . We need to check if this is a relative maximum by looking at the sign of the derivative before and after this point.
Verify Relative Maximum: We test the sign of the derivative around $x = 1.047$ . If the derivative changes from positive to negative at $x = 1.047$ , then it is a relative maximum.
Verify Relative Maximum: We test the sign of the derivative around $x = 1.047$ . If the derivative changes from positive to negative at $x = 1.047$ , then it is a relative maximum.Using the calculator, we find that the derivative is positive just before $x = 1.047$ and negative just after $x = 1.047$ . Therefore, $x = 1.047$ is a relative maximum.

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