#### To determine

**To find:**

the absolute maximum of the given function, and if possible, then its minimum too.

#### Answer

f(−5)=e45.

#### Explanation

**Given:**

f(x)=e10|x−2|−x2

**Formulae used:**

f'(x)=ddx(f(x))ddx(ex)=exddx(xn)=nxn−1ddx(u.v)=v.ddx(u)+uddx(v)

Consider f(x)=e10|x−2|−x2

Let us compute the first derivative of the function f(x)=e10|x−2|−x2 to identify the critical points.

For x>2

f(x)=e10x−20−x2f′(x)=(10−2x)e10x−20−x2

To find the critical points set *f’(x)=0* and solve for *x*.Thus f'(x)=0⇒(10−2x)e10x−20−x2=0⇒(10−2x)=0 (because exponential function is never zero)⇒x=5

Also notice that For x>5, 10-2x<0.Thus (10−2x)e10x−20−x2<0, for x>5.That is f'(x)<0,for x>5. So we see that *f(x)* is decreasing in (5,∞)

Similarly For 2<x<5, 10-2x>0.That is f'(x)>0,for 2<x>5. Thus we have that *f(x)* is increasing in (2,5) So to the right of 5 the function is decreasing whereas it is increasing to it’s left. So the point *x=5* is a local maximum.We may conclude this by computing second derivative at x=5 as well. Thus

f′(x)=(10−2x)e10x−20−x2f″(x)=(10−2x)2e10x−20−x2−2e10x−20−x2=e10x−20−x2[(10−2x)2−2]Now f''(5)=e10(5)−20−52[(10−2(5))2−2]=−2e5<0 Thus at *x=5,* the function has a local maximum.

Now the second case

For x<2

Calculate the derivative of the function to identify the critical points.

f(x)=e−10x+20−x2f′(x)=−(10+2x)e−10x+20−x2

To find the critical points set *f’(x)=0* and solve for *x*.Thus

(10+2x)e−10x+20−x2=0⇒x=−5

Notice that for x<−5, 10+2x<0

Thus (10+2x)e−10x+20−x2<0, for x<−5

i.e. if f'(x)<0, for x<−5

so we see that f(x) is decreasing in (−5,∞).

Similarly for −5<x<2, f'(x)>0.

Thus we have f(x) is increasing in (-5,2).

So to the left of -5 function is decreasing and to the right of -5 to 2, the function in increasing.

Hence at x=−5 there exist a local maximum.

We may conclude it by calculating second derivative at x=−5.

The second derivative of the function f(x)=e10|x−2|−x2 at x=−5

f′(x)=−(10+2x)e−10x+20−x2f″(x)=(10+2x)2e−10x+20−x2−2e−10x+20−x2=e−10x+20−x2[(10+2x)2−2]<0

Now consider x=2.

At x=2,

f'(x) is not defined.

Since limx→2−(f'(x))≠limx→2+(f'(x)).

Therefore x=2.

Plug the extreme point x=2 into e10|x−2|−x2

⇒y=1e4 i.e. is absolute minimum.

To check whether the function does not take any other smaller value.

Plug x=10 into e10|x−2|−x2.

⇒y=1e20 which is less than 1e4.

Therefore, the value of function f(x)=e10|x−2|−x2 at x=−5 is

f(−5)=e45

Sine f(−5) is bigger.

Thus, there only absolute maximum since f″(x)<0

**Conclusion:** Hence the value of the absolute maximum of the function f(x)=e10|x−2|−x2 is f(−5)=e45.