
How do you find the range of the function $f\left( x \right)={{x}^{2}}-25$ on the domain $-2\le x\le 3$?
Answer
554.1k+ views
Hint: Now we will differentiate the equation and find the nature of the graph in the given interval by first derivative test. Now we will find the values in values of the graph at critical points in the interval and hence find the maximum and minimum value of the function.
Hence we will get the range of the function.
Complete step-by-step solution:
Now first consider the given equation ${{x}^{2}}-25$
Now let us first differentiate the function. Hence we get the differentiation as 2x.
Now we know $2x>0,x>0$ and $2x<0,x<0$ and $2x=0$ for x = 0.
Hence by first derivative test we can say that the function is increasing for x > 0 and decreasing for x < 0.
Now since the function is defined on the interval $\left[ -2,3 \right]$ we will just consider this interval
Hence in the interval $[-2,0)$ the function is decreasing and in the interval $(0,3]$ the function is increasing and at x = 0 we have an extrema.
Now let us find $f\left( -2 \right),f\left( 0 \right)$ and $f\left( 3 \right)$ .
$f\left( -2 \right)={{\left( -2 \right)}^{2}}-25=4-25=-21$
$f\left( 0 \right)={{0}^{2}}-25=-25$
$f\left( 3 \right)=9-25=-16$
Now since the function has an extrema at x = 0 and $f\left( 3 \right)>f\left( 0 \right)$ we can say that the extrema is not maxima. Hence we have minima at x = 0. And hence $f\left( 0 \right)< f\left( x \right).......\left( 1 \right)$
Also we can say that the maximum value of function is attained at x = 3 for the interval $\left[ -2,3 \right]$ .
Hence we have for x belonging to interval $\left[ -2,3 \right]$, $f\left( 0 \right)< f\left( x \right)< f\left( 3 \right)$
Hence we get,
$\begin{align}
& \Rightarrow f\left( x \right)\in \left[ f\left( 0 \right),f\left( 3 \right) \right] \\
& \Rightarrow f\left( x \right)\in \left[ -25,-16 \right] \\
\end{align}$
Hence the range of the function is \[\left[ -25,-16 \right]\].
Note: Note that if we examine the function in the interval $\left[ -2,3 \right]$ we have the value of function as – 21 at x = - 2. Then the function decreases and reaches – 25 at x = 0. Then function starts increasing and reaches – 16 at x = 3. Hence we can draw a rough graph of the equation in the interval and determine maxima and minima of the function.
Hence we will get the range of the function.
Complete step-by-step solution:
Now first consider the given equation ${{x}^{2}}-25$
Now let us first differentiate the function. Hence we get the differentiation as 2x.
Now we know $2x>0,x>0$ and $2x<0,x<0$ and $2x=0$ for x = 0.
Hence by first derivative test we can say that the function is increasing for x > 0 and decreasing for x < 0.
Now since the function is defined on the interval $\left[ -2,3 \right]$ we will just consider this interval
Hence in the interval $[-2,0)$ the function is decreasing and in the interval $(0,3]$ the function is increasing and at x = 0 we have an extrema.
Now let us find $f\left( -2 \right),f\left( 0 \right)$ and $f\left( 3 \right)$ .
$f\left( -2 \right)={{\left( -2 \right)}^{2}}-25=4-25=-21$
$f\left( 0 \right)={{0}^{2}}-25=-25$
$f\left( 3 \right)=9-25=-16$
Now since the function has an extrema at x = 0 and $f\left( 3 \right)>f\left( 0 \right)$ we can say that the extrema is not maxima. Hence we have minima at x = 0. And hence $f\left( 0 \right)< f\left( x \right).......\left( 1 \right)$
Also we can say that the maximum value of function is attained at x = 3 for the interval $\left[ -2,3 \right]$ .
Hence we have for x belonging to interval $\left[ -2,3 \right]$, $f\left( 0 \right)< f\left( x \right)< f\left( 3 \right)$
Hence we get,
$\begin{align}
& \Rightarrow f\left( x \right)\in \left[ f\left( 0 \right),f\left( 3 \right) \right] \\
& \Rightarrow f\left( x \right)\in \left[ -25,-16 \right] \\
\end{align}$
Hence the range of the function is \[\left[ -25,-16 \right]\].
Note: Note that if we examine the function in the interval $\left[ -2,3 \right]$ we have the value of function as – 21 at x = - 2. Then the function decreases and reaches – 25 at x = 0. Then function starts increasing and reaches – 16 at x = 3. Hence we can draw a rough graph of the equation in the interval and determine maxima and minima of the function.
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