Question

If $f\left(x\right)=x^2-4x+5$ on $[0,3]$ then the absolute maximum value is:

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (D)

$5$

We have to find the absolute maximum value of $f\left(x\right)=x^2-4x+5$ on $[0,3]$.

Consider $f\left(x\right)=x^2-4x+5$

Since $f\left(x\right)$ is a polynomial it is continuous everywhere.

Differentiate $f\left(x\right)$ with respect to $x$ on both sides we get

$f^{\prime }\left(x\right)=2x-4$

Now to find the critical point equate $f^{\prime }\left(x\right)$ to $0$.

$\Rightarrow 2x-4=0$

$\Rightarrow 2x=4$

$\Rightarrow x=2$

The only critical point is $2$.

Now let us find the function value at the endpoints and the critical value to find the absolute maximum.

At $x=0$, $f\left(0\right)=0^2-4\left(0\right)+5=5$

At $x=2$, $f\left(2\right)=2^2-4\left(2\right)+5=4-8+5=1$

At $x=3$, $f\left(3\right)=3^2-4\left(3\right)+5=9-12+5=2$

From the above values we see that the absolute maximum is $5$ and it occur at $x=0$.