What is the extreme f(x)=xsqrt(x^2+25),-6<= x<= 4 ?

The extreme f(x)=xsqrt(x^2+25),-6<= x<= 4 is Minimum(-6,-6sqrt(61)),Maximum(4,4sqrt(41))

Popular >

extreme f(x)=xsqrt(x^2+25),-6<= x<= 4

extreme points

f (x) = x x^{2} + 25, - 6 \leq x \leq 4

Minimum (- 6, - 661), Maximum (4, 441)

Solution steps

Find the critical points:

x = - 6, x = 4

Domain of

x x^{2} + 25, - 6 \leq x \leq 4 : - 6 \leq x \leq 4

Find intervals:

Increasing

: - 6 < x < 4

Plug the extreme point

x = - 6

into

x x^{2} + 25 \Rightarrow y = - 661

Minimum (- 6, - 661)

Plug the extreme point

x = 4

into

x x^{2} + 25 \Rightarrow y = 441

Maximum (4, 441)

Minimum (- 6, - 661), Maximum (4, 441)

(- 2.3) (1)

f (x) = 1 - x + (\frac{( x ^{3} )}{3 + x})^{\frac{1}{2}}

f (x) = \frac{x ^{2} + 2 x - 3}{( x + 1 ) ^{2}}

a y = (x - 9) e^{x}

f (x, y) = x^{2} + y^{2} + 49

What is the extreme f(x)=xsqrt(x^2+25),-6<= x<= 4 ?
The extreme f(x)=xsqrt(x^2+25),-6<= x<= 4 is Minimum(-6,-6sqrt(61)),Maximum(4,4sqrt(41))