#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# SIR WILL YOU PLEASE DRAW AND EXPLAIN THE GRAPH OF X POWER X when x gr8r than 1 it is easy but when x is less than 1 it is becoming a bit complicated please draw it and explain 10 years ago

Dear student,

For positive x, it will start near 1, for x near zero, then will dip down before increasing up to f(1) = 1, then it will increase more than exponentially for x > 1.

For negative values of x, most values of f(x) will be complex numbers, so you would only be able to plot scattered points on a real number graph.
For example (-1/2)^(-1/2) = 1/sqrt(-1/2) which is imaginary, but (-1/3)^(-1/3) is 1/cube_root(-1/3) which is a real number. Rational negative values of x should produce either pure real or pure imaginary values of f(x). Irrational negative values of x will produce f(x) being a complex number.

All negative integer values of x will result in f(x) real. As x becomes more negative, the values of f(x) will be exponentially closer to zero, but jumping between positive and negative values. Example (-2)^(-2) = (-1/2)^2 = 1/4, but (-3)^(-3) = (-1/3)^3 = -1/27
If you could plot an additional imaginary axis (perpendicular to the x and to the y axis, in 3-dimensions), f(x) would be seen spiraling around the x-axis, approaching the x-axis but never quite getting there, as x gets more negative. For values of x where f(x) is real, this is where the spiral intersects the 'real' x-y plane.