Home > Calculus > Interpreting the Sign of the First Derivative (Increasing and Decreasing Functions)

Is #f(x)=e^xsqrt(x^2-x)# increasing or decreasing at #x=3#?

Answer 1

Ezra Adams

#f(x)=e^xsqrt(x^2-x)# is increasing at #x=3#

A function #f(x)# is increasing at #x=a# if #f'(a)>0# and is decreasing at #x=a# if #f'(a)<0#.

As #f(x)=e^xsqrt(x^2-x)#, using product rule

#f'(x)=e^xsqrt(x^2-x)+e^x xx1/(2sqrt(x^2-x))xx(2x-1)#

= #e^xsqrt(x^2-x)+(e^x(2x-1))/(2sqrt(x^2-x))#

and at #x=3#

#f'(x)=e^3sqrt(3^2-3)+(e^3(2xx3-1))/(2sqrt(3^2-3))#

= #e^3sqrt6+(5e^3)/(2sqrt6)=e^3sqrt6(1+5/12)=e^3sqrt6xx17/12#

which is clearly positive.

Hence, #f(x)=e^xsqrt(x^2-x)# is increasing at #x=3#.

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Answer 2

Elijah Abram

To determine if ( f(x) = e^x \sqrt{x^2 - x} ) is increasing or decreasing at ( x = 3 ), we need to examine the sign of its derivative at that point. The derivative of ( f(x) ) is ( f'(x) = e^x \sqrt{x^2 - x} + \frac{e^x (2x - 1)}{2 \sqrt{x^2 - x}} ).

Plugging in ( x = 3 ) gives ( f'(3) = e^3 \sqrt{3^2 - 3} + \frac{e^3 (2 \cdot 3 - 1)}{2 \sqrt{3^2 - 3}} ).

Since both terms in the derivative are positive (as ( e^3 ) is positive and the square roots are positive for ( x > 1 )), ( f'(3) ) is positive. Therefore, ( f(x) ) is increasing at ( x = 3 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.