11. Consider a function y = x⁵ – 4, then the slope of y vs. x curve will be
a. always positive
b. linearly increasing with increasing x value
c. always an integer
d. increases logarithmically

Slope of the Curve y = x⁵ − 4

The slope of the curve y = x⁵ − 4 is given by its derivative:

dy/dx = 5x⁴

This derivative equals zero at x = 0 and remains non-negative everywhere since x⁴ ≥ 0 for all real x.
Hence, the slope is always positive except at the origin. Thus, option (a) always positive is correct.

Option Analysis

a. always positive: Correct. The derivative 5x⁴ satisfies 5x⁴ ≥ 0 for all real x, with equality only at x = 0.
Example values like 5(−2)⁴ = 80 > 0 and 5(2)⁴ = 80 > 0 confirm positivity across the domain.

b. linearly increasing with increasing x value: Incorrect. While 5x⁴ rises for x > 0
(e.g., from 5 at x = 1 to 80 at x = 2), linear regression on slopes over
x ∈ [−3, 3] yields r = 0 due to symmetry and quartic acceleration, not straight-line growth.

c. always an integer: Incorrect. Slopes like 5(1)⁴ = 5 (integer) contrast with
5(0.5)⁴ = 2.5 (non-integer), proving non-integer values exist.

d. increases logarithmically: Incorrect. Logarithmic growth is concave down (e.g., ln x),
but 5x⁴ shows convex acceleration (e.g., 5 → 80 → 405 from x = 1 to 2 to 3), mismatched by shape.

Derivative Proof

Differentiating y = x⁵ − 4 using the power rule:

dy/dx = 5x⁴ − 0 = 5x⁴

Since x⁴ is even and non-negative, multiplying by positive 5 ensures dy/dx ≥ 0
with the minimum at x = 0. This quintic curve rises symmetrically and steepens outward.

Common exam distractors include linear and logarithmic confusion with quartic derivatives.
Verifying through limits: as |x| → ∞, slope → ∞ faster than linear.