QuestionIdentify the true statement about the function $f(x)=(x-3)(x-7)^{2}$ from the options given.

Studdy Solution

STEP 1

Assumptions1. We are given a function $f(x)=(x-3)(x-7)^{}$ . . We need to determine which of the given options is true regarding the zeros and relative extrema of the function.

STEP 2

First, let's find the zeros of the function. The zeros of a function are the x-values for which the function equals zero. We can find them by setting the function equal to zero and solving for x.
$f(x) =0$ $(x-)(x-7)^{2} =0$

STEP 3

Now, solve the equation for x. A product equals zero if and only if at least one of the factors equals zero. So, we can set each factor equal to zero and solve for x.
$x-3 =0 \quad or \quad (x-7)^{2} =0$

STEP 4

olve the equations for x.
$x =3 \quad or \quad x =7$ So, the zeros of the function are $x =3$ and $x =7$ .

STEP 5

Now, let's find the relative extrema of the function. The relative extrema of a function are the local minimum and maximum points. We can find them by taking the derivative of the function and setting it equal to zero.
The derivative of the function is $f'(x) = (x-7)^{2} +2(x-3)(x-7)$

STEP 6

Set the derivative equal to zero and solve for x.
$f'(x) =0$ $(x-)^{2} +2(x-3)(x-) =0$

STEP 7

olve the equation for x.
The solutions are $x =3$ and $x =7$ .

STEP 8

Now, let's determine whether these points are relative minima or maxima. We can do this by using the second derivative test. The second derivative of the function is $f''(x) =2(x-7) +2(x-3)$

STEP 9

Evaluate the second derivative at $x =3$ and $x =7$ .
$f''(3) =2(3-7) +2(3-3) = -8$ $f''(7) =2(7-7) +2(7-3) =8$

STEP 10

According to the second derivative test, if $f''(x) >0$ , then $f(x)$ has a relative minimum at that point. If $f''(x) <0$ , then $f(x)$ has a relative maximum at that point.
So, $(3,0)$ is a zero and a relative maximum, and $(7,0)$ is a zero and a relative minimum.
Therefore, the correct answer is option b. " $(3,0)$ is a zero and a relative maximum."

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